Proof of Lemma 3.2 (ACA $_{0}$ )

Let $G=\left \langle V,E\right \rangle $ be a locally finite graph with arbitrarily many E-disjoint rays. Consider the line graph $L(G)$ of G, i.e., the graph whose vertices are the edges of G and whose edges are the $ ((x,y),(y,z))$ for $(x,y)\neq (y,z)\in E$ . As G is locally finite so is $ L(G)$ . (The only way a given $(u,v)$ can have an $(x,y)$ as a neighbor is if they have a vertex in common. So if $(u,v)$ had infinitely many neighbors, one of u or v would also have such in G.)

A set of k many E-disjoint rays $R_{i}=\left \langle V_{i},E_{i}\right \rangle $ in G with isomorphisms $f_{i}$ produces k many V-disjoint rays $\bar {R}_{i}$ with vertices $(x_{i,n},x_{i,n+1})$ in $ L(G)$ where we write $x_{i,n}$ for $f_{i}(n)$ . Applying the hypothesis of the lemma gives us infinitely many V-disjoint rays $T_{i}$ with vertices $ (x_{i,j},x_{i,j+1})$ in $L(G)$ .

Now we use the local finiteness of G to construct the required infinitely many E-disjoint rays $Q_{i}$ in $G.$ Fix i and $T=T_{i}$ and $ x_{j}=x_{i,j} $ . By the local finiteness of G, for every $v\in V$ there are only finitely many n such that $v\in \{x_{n},x_{n+1}\}.$ (Otherwise, say v is $x_{n}$ for infinitely many n and then $(v,x_{n+1})$ is a vertex in the $L(G)$ ray for all of these n. That means, however, that these edges are all distinct and so v has infinitely many neighbors in G for a contradiction.)

We build $Q=Q_{i}$ by recursion starting with $a_{0}=x_{0}$ and let $n_{0}$ be the largest n such that $a_{0}=x_{n}$ . Let $a_{1}=x_{n_{0}+1}$ so $ (a_{0},a_{1})\in E$ . Inductively, take $n_{k+1}$ ( $>n_{k}$ ) the largest n such that $a_{k}=x_{n}$ and set $a_{k+1}=x_{n_{k}+1}$ . (We can find this n by ACA $_{0}$ .) This recursion produces sequences $n_{k},a_{k+1}$ with $ Q=\left \langle a_{k}\right \rangle $ a ray in G and $(a_{n},a_{n+1})\in T$ for every n. Let this Q be $Q_{i}$ .

Claim: The $Q_{i}$ are E-disjoint in G as required. If not, we have $ (a_{i,n},a_{i,n+1})=(a_{j,m},a_{j,m+1})$ for some $i\neq j$ and n and m. However, $(a_{i,n},a_{i,n+1})\in T_{i}$ and $(a_{j,m},a_{j,m+1})\in T_{j}$ contradicting the V-disjointness of the $T_{i}$ in $L(G)$ .

Proof of Lemma 3.1 ( $\Sigma _{1}^{1}$ -AC $_{0}$ )

We are given a graph $G=\left \langle V,E\right \rangle $ with arbitrarily many E-disjoint rays. By $\Sigma _{1}^{1}$ -AC $_{0}$ choose a sequence $ S_{k}=\left \langle R_{k,1},\ldots ,R_{k,k}\right \rangle $ which consists, for each k, of k many E-disjoint rays in G. Now we construct the desired subgraph $G^{\prime }$ of $G.$ It has the same set of vertices $ V=\{v_{i}|i\in N\}$ as G. We specify its edges by providing a recursive construction of sets $E_{i}$ of edges putting in a set of edges at each step. We guarantee that each $E_{i}$ is a union of finitely many finite sets of E-disjoint rays in G and that after stage s no edge with a vertex $ v_{i}$ for $i<s$ is ever put into E after stage s.

Begin at stage $0$ by putting all the edges in $R_{1,1}$ into $E_{1}$ . Proceeding recursively at stage k we have $E_{k}$ and consider $ S_{k}=\left \langle R_{k,1},\ldots ,R_{k,k}\right \rangle $ . Each $v_{j}$ for $ j<k$ appears in $R_{k,i}$ at most once for each $i<k$ as $R_{k,i}$ is a ray. As we have the whole sequence of the $S_{k}$ we can find (using only RCA $ _{0} $ ) the $i,j$ such that $v_{j}\in R_{k,i}$ and the location n in the sequence where it occurs, say as $x_{k,i,n_{j}}$ . We now put into $E_{k+1}$ all the edges appearing in any $R_{k,i}$ after all the $x_{k,i,n_{j}}$ for $ j<k$ which are defined. Let $E^{\prime }=\cup E_{k}$ . As, for every k, we have put in a tail of each $R_{k,i}$ for $i<k$ into E $_{k}$ we have guaranteed that $G^{\prime }=\left \langle V,E^{\prime }\right \rangle $ contains arbitrarily many E-disjoint rays.

Thus we only need to show that $G^{\prime }$ is locally finite. Consider any vertex $v_{k}$ . No edge containing $v_{k}$ as a vertex is put in after stage k. On the other hand, $E_{k}$ is the union of finitely many finite sets of E-disjoint rays (all of which have been computed uniformly). Each set of E-disjoint rays in this union has $v_{k}$ appearing at most once in each of its rays. Thus at most two edges containing $v_{k}$ appear in each of the finitely many rays in this set. Thus there are only finitely many edges containing $v_{k}$ in each of the finite sets of E-disjoint making up $E_{k}$ . All in all, this makes only finitely many edges containing $v_{k}$ get put into $G^{\prime }$ . (In fact, we can compute the number of such edges in $ \mathcal {N}$ .)

Proof Let $G=\left \langle V,E\right \rangle $ and $V=\{v_{i}|i\in N\}$ . For each n and n-tuple $\left \langle v_{i_{j}}|j<n\right \rangle $ of distinct vertices of G consider the tree $T_{i,n}$ whose nodes are n-tuples of disjoint paths in G all of the length the height of the node in $T_{i,n}$ . The root of $T_{i,n}$ is $\left \langle v_{i_{j}}|j<n\right \rangle $ . If $\mu =\left \langle \sigma _{j}|j<n\right \rangle \in T_{i,n}$ then its immediate successors are all $\nu =\left \langle \tau _{j}|j<n\right \rangle $ such that for each j, $\tau _{j}$ is an extension of $\sigma _{j}$ by one of the finitely many vertices v such that there is an edge from the last vertex in $\sigma _{j}$ to v and the $\tau _{j}$ are disjoint paths in G. As G is locally finite, the $T_{i,n}$ are finitely branching trees (recursive in $ G^{\prime }$ ). Thus by ACA $_{0}$ (recursively in a few jumps of G) we can get the set of $\left \langle i,n\right \rangle $ such that $T_{i,n}$ has a branch and, indeed, a sequence $S_{i,n}$ each a branch in $T_{i,n}$ for each such $\left \langle i,n\right \rangle $ . Every such branch provides a subgraph $G_{i,n}$ of G which consists of n disjoint rays. We can now just take the desired $H_{n}$ to be $G_{i,n}$ for the least i such that $T_{i,n}$ has a branch.

Proof To prove the implication from left to right, suppose G is a locally finite X-graph with arbitrarily many Y-disjoint rays. By Proposition 3.3 and BGS [Theorem 8.2], G has infinitely many Y-disjoint rays. For the other direction, note that the graphs used in BGS [Theorems 4.1, 4.2, and 5.9] to deduce ACA $_{0}$ from IRT $_{XY}$ are disjoint unions of finitely branching trees and so locally finite.

Proof Suppose we are given a graph H with arbitrarily many disjoint rays. Let $ H^{\prime }$ be a locally finite subgraph with arbitrarily many rays. Again, Proposition 3.3 and Theorem 8.2 of BGS give us IRT $_{XY}$ .

In fact, over ACA $_{0}$ we have equivalences.

Proof We are given a graph G with arbitrarily many disjoint rays and want to build a locally finite subgraph with the same property. We begin with the subgraph of the given graph consisting of infinitely many disjoint rays asserted to exist by IRT $_{XY}$ . As IRT $_{XY}\rightarrow $ ACA $_{0}$ (BGS [Theorem 5.1]), we can use ACA $_{0}$ to thin out this subgraph so that our new nth ray is simply the nth given ray above the last time any vertex less than n appears in it. (Any vertex appears at most once in any ray.) Thus every vertex less than n appears in at most n many of these new rays. In each one it has edges to at most two other vertices. Thus it is a locally finite subgraph of the original graph and also contains infinitely many disjoint rays. The equivalence now follows from Proposition 3.6.

Proof Each IRT $_{XY}$ is a THA by BGS [Theorem 5.1] and so we have the result by Proposition 3.7.

Notation 4.3. If $\mathcal {G}$ is a filter on a t-forcing $\mathcal {P}$ which is generic for a class $\mathcal {D}$ of dense sets containing at least the $D_{n}=\{\left \langle \tau ,T\right \rangle ||\tau |\geq n\}$ , then the generic function $G:N\mathbb {\rightarrow }N$ associated with $\mathcal {G}$ is $\cup \{\tau |\exists T(\left \langle \tau ,T\right \rangle \in \mathcal {G\} }$ . We then say that G is $\mathcal {D}$ -generic on $\mathcal {P}$ . (Note that this G is always a function from N to N by the definitions of $ \mathcal {P}$ being a t-forcing and of $\mathcal {G}$ being a $\mathcal {D}$ -generic filter on $\mathcal {P}$ .) We also say that a $G:N\rightarrow N$ is on $\left \langle \tau ,T\right \rangle $ if $G\in \lbrack T]$ , i.e., $\forall n(G\upharpoonright n\in T)$ . So if G is $\{D_{n}\}$ -generic it is on every $\left \langle \tau ,T\right \rangle \in \mathcal {G}$ .

We denote by $\mathcal {N}[G]$ the structure for second-order arithmetic with first order part the same as $\mathcal {N}$ (i.e., N) and second order part the closure of $S(\mathcal {N})\cup \{G\}$ under $\Delta _{1}^{0}$ -CA.

For $\sigma \in T$ , $T^{\sigma }=\{\rho \in T| \ \sigma \supseteq \rho \vee \rho \supseteq \sigma \}$ is what we call the tree T above $ \sigma $ .

Proof The third clause follows immediately from the second by the countability of $ S(\mathcal {N})$ and then by taking $R_{i}=\{\rho \in N^{<N}|\rho \subset C_{i}\}$ . We prove each of the first two assertions by specifying the appropriate collections of dense sets.

1. By a classic result of Friedman [Reference Friedman8], it suffices to show that for any $\Sigma _{1}^{0}$ formula $\exists k\Phi (k,m,A,G)$ with $A\in S( \mathcal {N})$ such that $\mathcal {N}[G]\vDash \exists k\Phi (k,m,A,G)$ there is an $\mathcal {N}$ -least $m^{\prime }$ such that $\mathcal {N}[G]\vDash \exists k\Phi (k,m^{\prime },A,G)$ . By the definition of G there is a condition $\left \langle \tau ,T\right \rangle \in \mathcal {G}$ such that $ \tau \Vdash \exists k\Phi (k,m,A,G)$ . We show that the conditions that guarantee that there is an $\mathcal {N}$ -minimal such m are dense below $ \left \langle \tau ,T\right \rangle $ and so we can extend $\mathcal {D}$ to guarantee that $\mathcal {N}[G]\vDash $ RCA $_{0}$ as there are only countably many $\Sigma _{1}^{0}$ formulas $\exists k\Phi (k,m,A,G)$ and conditions $ \left \langle \tau ,T\right \rangle $ . As $\mathcal {P}$ is an et-forcing the set $\{m^{\prime }\leq m|(\exists \tau ^{\prime }\in Ext(\left \langle \tau ,T\right \rangle )(\tau ^{\prime }\Vdash \exists k\Phi (k,m^{\prime },A,G))\}$ is $\mathbf {\Sigma }_{1}^{0}$ in $\mathcal {N}$ and so as $\mathcal {N}\vDash $ RCA $_{0}$ , there is an $\mathcal {N}$ -least such $m^{\prime }$ with an associated $\left \langle \tau ^{\prime },T^{\prime }\right \rangle $ . If $ \mathcal {D}$ includes the corresponding dense set consisting of such $ \left \langle \tau ^{\prime },T^{\prime }\right \rangle $ for each $ \left \langle \tau ,T\right \rangle $ , it is clear that $m^{\prime }$ is the least m such that $\mathcal {N}[G^{\prime }]\vDash \exists k\Phi (k,m,A,G^{\prime })$ for any $\mathcal {D}$ -generic $G^{\prime }$ as desired. (Otherwise, there would be a $\left \langle \tau ^{\prime \prime },T^{\prime \prime }\right \rangle \leq \left \langle \tau ^{\prime },T^{\prime }\right \rangle , $ a $k^{\prime }\in N$ , and an $m^{\prime \prime }<_{\mathcal {N }}m^{\prime }$ such that $\tau ^{\prime \prime }\Vdash \Phi (k^{\prime },m^{\prime \prime },A,G)$ .)

2. Consider any R as in the claim. We again want to specify the additional dense sets needed. Consider an arbitrary function in some $\mathcal {N}[G]$ . By the definition of $\mathcal {N}[G]$ (and basic facts of about recursive functions true in RCA $_{0}$ ), it is of the form $\Phi _{e}^{A\oplus G}$ for some $e\in N$ and $A\in S(\mathcal {N})$ .

If for every $\left \langle \tau ,T\right \rangle \in \mathcal {G}$ there is a $ \rho \notin R$ ( $\rho \in N^{<N}$ ) and a $\tau ^{\prime }\in Ext(\left \langle \tau ,T\right \rangle )$ such that $\tau ^{\prime }\Vdash \Phi _{e}^{A\oplus G}(l)=\rho (l)$ for every $l<|\rho |$ then the set of conditions guaranteeing that $\Phi _{e}^{A\oplus G}$ is not on R is dense. Thus we may assume that we have a $\left \langle \tau ,T\right \rangle \in \mathcal {G}$ such that for every $\tau ^{\prime }\in Ext(\left \langle \tau ,T\right \rangle )$ and every $\rho $ , if $\tau ^{\prime }\Vdash \Phi _{e}^{A\oplus G}(l)=\rho (l)$ for every $l<|\rho |$ , then $\rho \in R$ .

Next, if there is a $\tau ^{\prime }\in Ext(\left \langle \tau ,T\right \rangle )$ and an $l\in N$ such that there is no $m\in N$ and $\tau ^{\prime \prime }\in Ext(\left \langle \tau ,T\right \rangle )$ with $\tau ^{\prime \prime }\supseteq \tau ^{\prime }$ such that $\tau ^{\prime \prime }\Vdash \Phi _{e}^{A\oplus G}(l)=m$ then the associated $\left \langle \tau ^{\prime },T^{\prime }\right \rangle \leq \left \langle \tau ,T\right \rangle $ guarantees that $\Phi _{e}^{A\oplus G}(l)\uparrow $ and so $\Phi _{e}^{A\oplus G}$ is again not a branch on R. Thus we may assume that for every $l\in N$ and $\tau ^{\prime }\in Ext(\left \langle \tau ,T\right \rangle )$ there is an $m\in N$ and $\tau ^{\prime \prime }\in Ext(\left \langle \tau ,T\right \rangle )$ with $\tau ^{\prime \prime }\supseteq \tau ^{\prime }$ such that $\tau ^{\prime \prime }\Vdash \Phi _{e}^{A\oplus G}(l)=m$ .

We now prove that there is a branch f on R which is in $S(\mathcal {N})$ for a contradiction: By our last assumption on $\left \langle \tau ,T\right \rangle $ , we can define an $f:N\mathbb {\rightarrow }N$ by recursion in $\mathcal {N}$ starting with our $\left \langle \tau ,T\right \rangle $ and $ \tau _{-1}=\tau $ : We build sequences of $\tau _{l}\in Ext(\left \langle \tau ,T\right \rangle )$ and $m_{l}$ such that $\tau _{l}\Vdash \Phi _{e}^{A\oplus G}(l)=m_{l}$ and $\tau _{l}\subseteq \tau _{l+1}$ . This is a recursive procedure in $\mathcal {N}$ as $\mathcal {P}$ is an et-forcing and so we can search for the next witnesses (and find them) effectively in $\mathcal {N}$ . By our first assumption on $\left \langle \tau ,T\right \rangle $ , the sequence $\left \langle m_{l}|l<n\right \rangle $ is in R for every $n\in N$ and so f is the desired branch on R in $\mathcal {N}$ .

Proof Let $C_{i}$ list the subsets of N which are in $\mathcal {N}_{0}$ but not in $\mathcal {N}$ . Apply Theorem 4.5.3 to get the desired collection of dense sets.

Proof We begin the plan outlined above with quantifier free G- $\Pi _{1}^{1}$ sentences $\Lambda $ with constants from N and $S(\mathcal {N})$ . Here the truth of both $\Lambda $ and $\lnot \Lambda $ are preserved from $\mathcal {N} $ to $\mathcal {N}_{\infty }$ as $N=N_{\infty }$ and $S(\mathcal {N})\subseteq S(\mathcal {N}_{\infty })$ . Suppose next that $\Lambda =\Delta _{0}\wedge \Delta _{1}$ . As $\mathcal {N}\vDash \lnot \Lambda $ , $\mathcal {N}\vDash \lnot \Delta _{i}$ for at least one $i\in \{0,1\}$ . By induction then $ \mathcal {N}_{\infty }\vDash \lnot \Delta _{i}$ as well as required. If $ \Lambda =\Delta _{0}\vee \Delta _{1}$ and $\mathcal {N}\vDash \lnot \Lambda $ then $\mathcal {N}\vDash \lnot \Delta _{0}\wedge \lnot \Delta _{1}$ and so by induction $\mathcal {N}_{\infty }\vDash \lnot \Delta _{0}\wedge \lnot \Delta _{1}$ and $\mathcal {N}_{\infty }\vDash \lnot (\Delta _{0}\vee \Delta _{1})$ as required. Next, suppose $\Lambda =\forall x\Delta (x)$ and $\mathcal {N} \vDash \lnot \Lambda $ . Choose an $n\in N$ such that $\mathcal {N}\vDash \lnot \Delta (n)$ . By induction, $\mathcal {N}_{\infty }\vDash \lnot \Delta (n)$ and so $\mathcal {N}_{\infty }\vDash \lnot \forall x\Delta (x)$ . Suppose $\Lambda =\exists x\Delta (x)$ and $\mathcal {N}\vDash \lnot \Lambda $ . If $ \mathcal {N}_{\infty }\vDash \exists x\Delta (x)$ choose a witness $n\in N_{\infty }=N$ so that $\mathcal {N}_{\infty }\vDash \Delta (n)$ . As $\Delta (n)$ is also a sentence in $\Gamma $ , we have that $\mathcal {N}\vDash \Delta (n)$ by induction for the desired contradiction. Finally, if $\Lambda =\forall X\Delta (X)$ and $\mathcal {N}\vDash \lnot \Lambda $ , choose a $W\in S(\mathcal {N})$ such that $\mathcal {N}\vDash \lnot \Delta (W)$ . As $\Delta (W)$ is a sentence in $\Gamma $ , we again have $\mathcal {N}_{\infty }\vDash \lnot \Delta (W)$ as required.

Proof The argument for quantifier free sentences is as in Theorem 4.11 as are the inductive cases for $\wedge $ , $\vee $ , $\exists x$ , $\forall x$ , and $\forall X$ . Here we also have to begin with sentences $ \Lambda $ of the form $\exists Y\Theta (\bar {x},\bar {y},Y)$ with $\Theta $ a $\Sigma _{3}^{0}$ formula with constants from N and $S(\mathcal {N})$ and suppose that $\mathcal {N}\vDash \lnot \Lambda $ . As in [Reference Hirschfeldt, Shore and Slaman15, last paragraph of page 5818], the point here is that for any model $\mathcal {N}$ of RCA $_{0}$ the failure of a sentence $\exists Y\Theta (Y)$ with $\Theta $ being $\Sigma _{3}^{0}$ (with set constants $\bar {W}$ ) is equivalent to their being, for each $k\in N$ , a specifically defined tree $T_{k}$ (recursive in $\bar {W}$ ) which has no branch in the model. Thus none of these trees has a branch in $\mathcal {N}$ and so by our assumptions on the forcings none in $\mathcal {N}_{\infty }$ either. Thus $\mathcal {N} _{\omega }\vDash \lnot \exists Y\Theta (\bar {x},\bar {y}.Y)$ as required.

Proof As there are here no conservation results to verify the proof is simply the basic argument given above for the construction. Then one simply notes that by the choice of forcings no branch is added on for any $R_{i}$ at any successor step and so none enter at a limit level either.

Proof The set of vertices of $H^{\prime }$ is just that of H which, without loss of generality, we may take to be N. Thus we only need to specify the edges of $H^{\prime }$ . We use a pairing function to view the numbers n as all possible edges. We write $M(v,n)$ to mean that the vertex v is an element of the edge n.

Our conditions $\left \langle \tau ,T\right \rangle $ will satisfy various requirements in addition to the ones common to all et-forcings. The first is that, the trees are binary (i.e., subsets of $2^{<N}$ ) and for any $\sigma \in T$ , if (the pair) n is not an edge in H then $\sigma (n)=0$ . Thus each generic $G:N\mathbb {\rightarrow }N$ can be seen as a subgraph of H in $\mathcal {N}[G]$ . (G supplies the characteristic function of the set of edges. The set of vertices we have already set to be N.)

The intuition behind the rest of the definition of the notion of forcing is that we want to be able, on one hand, to specify that the set of edges from some sequence of disjoint rays of length m can be added to the final graph. On the other, for an arbitrary vertex v we want to be able to specify that no additional edges containing v can be added to the graph and to guarantee that only finitely many of the edges already guaranteed to be in the graph have v as a vertex.

To these ends, for each potential condition $\left \langle \tau ,T\right \rangle $ we first specify various sets of edges n based on what the condition says about their membership in G. First we have the $n>|\tau |$ such that all branches G in T have $G(n)=1$ . We denote the set of such n as $Y^{\left \langle \tau ,T\right \rangle }=\{n>|\tau ||(\forall \sigma \in T)(\sigma (n)=1)\}$ . Next we have the ones with $G(n)=0$ for all $ G$ on T, $N^{\left \langle \tau ,T\right \rangle }=\{n>|\tau ||(\forall \sigma \in T)(\sigma (n)=0)\}$ . Finally, we have the n at which G can go either way, . Note that as T is binary all these sets are in $ \mathcal {N}$ (even uniformly in $\left \langle \tau ,T\right \rangle $ ). We require that every $n>|\tau |$ is in one of these (clearly disjoint) sets.

We also impose some requirements on the nature of the first two of these three sets. For $Y^{\left \langle \tau ,T\right \rangle }$ there is a function $f_{\left \langle \tau ,T\right \rangle }=f$ such that $\forall v(f(v)=|\{n\in Y^{\left \langle \tau ,T\right \rangle }|M(v,n)\}|)$ . So not only are there only finitely many $n\in Y^{\left \langle \tau ,T\right \rangle }$ such that $ v $ is a member of the edge n but we know how many and so we can uniformly determine which they are. Finally, for $N^{\left \langle \tau ,T\right \rangle }$ there is a finite set $A_{\left \langle \tau ,T\right \rangle }=A$ such that $(\forall n\in N^{\left \langle \tau ,T\right \rangle })(\exists v\in A)M(n,v)$ and $(\forall n>|\tau |)(\forall v\in A)(M(n,v)\rightarrow n\notin U^{\left \langle \tau ,T\right \rangle })$ . We call such f and A witnesses that a $\left \langle \tau ,T\right \rangle $ satisfying the other requirements on $Y^{\left \langle \tau ,T\right \rangle }$ , $N^{\left \langle \tau ,T\right \rangle }$ and $U^{\left \langle \tau ,T\right \rangle }$ is a condition. Note that f is uniquely determined but A need not be.

Given this set of conditions, the forcing partial order is just the one defined by the basic requirement for et-forcings in Definition 4.1.2.

Note that if $\left \langle \tau ,T\right \rangle $ is a condition with witnesses f and A and $\sigma \in T$ with $|\sigma |>|\tau |$ then $ \left \langle \sigma ,T^{\sigma }\right \rangle $ (as defined in Notation 4.3) is also a condition with the same A and a slightly modified $ f^{\prime }$ as witnesses: let $f^{\prime }(v)=f(v)-|\{n\in Y^{\left \langle \tau ,T\right \rangle }|~|\tau |\leq n<|\sigma |~\wedge M(v,n)\}|$ . Thus we have defined an et-forcing.

We now argue that we can describe a collection $\mathcal {D}$ of dense sets that guarantee that any $\mathcal {D}$ -generic G determines (as described above) a locally finite subgraph $H^{\prime }$ of H which contains arbitrarily many disjoint rays.

First, we claim that for each vertex v and condition $\left \langle \tau ,T\right \rangle $ with witnesses A and f as above there is a $ \left \langle \tau ^{\prime },T^{\prime }\right \rangle \leq \left \langle \tau ,T\right \rangle $ with witnesses $f^{\prime }$ and $A^{\prime }$ such that $ v\in A^{\prime }$ . In fact, it is easy to see from the definition of the allowed conditions that for any $v\notin A_{\left \langle \tau ,T\right \rangle }$ , there is a $T^{\prime }\subseteq T$ such that $ \left \langle \tau ,T^{\prime }\right \rangle $ is a condition with witness f and $A^{\prime }=A\cup \{v\}$ . We refine T by removing any $\sigma \supset \tau $ such that there is an $n\geq |\tau |$ which is an edge containing v as a vertex such that both and are in T and $\sigma (n)=1$ . This moves n from $U^{\left \langle \tau ,T\right \rangle }$ to $N^{\left \langle \tau ,T\right \rangle }$ and so does not affect the calculation of the required f.

So, by genericity (for these dense sets), for any vertex v there is a condition $\left \langle \tau ,T\right \rangle $ in the generic filter such that it has a witness $A_{\left \langle \tau ,T\right \rangle }$ containing v .

As for this enforcing local finiteness, consider any condition $\left \langle \tau ,T\right \rangle $ with witnesses such that $v\in A_{\left \langle \tau ,T\right \rangle }$ . We claim that for any G on T there are only finitely many n with $G(n)=1$ and $M(v,n)$ . Of course, there are at most $|\tau |$ many edges $n<|\tau |$ such that $M(v,n)$ . There are only $f(v)$ many $n\in Y^{\left \langle \tau ,T\right \rangle }=Y^{\left \langle \tau ,T^{\prime }\right \rangle }$ such that $M(v,n)$ and by our second condition about $v\in A$ there are no $n\in U^{\left \langle \tau ,T\right \rangle }$ such that $ M(v,n)$ . Of course, there are no $n\in N^{\left \langle \tau ,T^{\prime }\right \rangle }$ such that $G(n)=1$ . (We have, in fact, uniformly over all $ G$ on T calculated the number of n such that $M(v,n)$ .)

Finally, we show that for each $m\in N$ and condition $\left \langle \tau ,T\right \rangle $ with witnesses as above, there is an extension $ \left \langle \tau ,T^{\prime }\right \rangle $ such that for any G on $ \left \langle \tau ,T^{\prime }\right \rangle $ the associated graph contains $ m$ many disjoint rays.

By assumption there is a sequence $\left \langle V_{i},E_{i},f_{i}\right \rangle _{i<m+|\tau |}$ of disjoint rays in H. Obviously, there are at most $|\tau |$ many n such that $\tau (n)=0$ . As each edge n can appear in at most one $E_{i}$ we can thin out the given sequence to one $\left \langle V_{i}^{\prime },E_{i}^{\prime },f_{i}^{\prime }\right \rangle _{i<m}$ of disjoint rays none of which contains an edge n for which $\tau (n)=0$ .

Next we deal with the conditions imposed by $A_{\left \langle \tau ,T\right \rangle }$ . For each $i<m$ we let $g(i)=0$ if no $v\in A_{\left \langle \tau ,T\right \rangle }$ is in $V_{i}^{\prime }$ . Otherwise we let $g(i)=\max \{n|f_{i}^{\prime }(n)\in A_{\left \langle \tau ,T\right \rangle }$ and $f_{i}^{\prime }(n)\in V_{i}^{\prime }\}+1$ . Our construction so far guarantees that $g:m\rightarrow N$ is a member of $ \mathcal {N}$ . Thus we can thin out $\left \langle V_{i}^{\prime },E_{i}^{\prime },f_{i}^{\prime }\right \rangle _{i<m}$ by taking the tail of each $V_{i}^{\prime }$ beyond $g(i)$ to get $V_{i}^{\prime \prime }=V_{i}^{\prime }-\{f_{i}^{\prime }(n)|n<g(i)\}$ , $E_{i}^{\prime \prime }=E_{i}^{\prime }-\{\left \langle f_{i}^{\prime }(n),f_{i}^{\prime }(n+1)\right \rangle |n<g(i)\}$ , and $f_{i}^{\prime \prime }(n)=f_{i}^{\prime }(n+g(i))$ . Let $E^{\prime \prime }=\cup \{E_{i}^{\prime \prime }|i<m\}$ . This sequence clearly provides m many disjoint rays in H. Thus (by genericity) it suffices to define a condition $\left \langle \tau ,T^{\prime }\right \rangle \leq \left \langle \tau ,T\right \rangle $ such that $E^{\prime \prime }\subseteq Y^{\left \langle \tau ,T^{\prime }\right \rangle }$ as then for any G on $\left \langle \tau ,T^{\prime }\right \rangle $ , $G(n)=1$ for all the edges $n\in \ E_{i}^{\prime \prime }$ for every $i<m$ and so $ \left \langle V_{i}^{\prime \prime },E_{i}^{\prime \prime },f_{i}^{\prime \prime }\right \rangle _{i<m}$ shows in $\mathcal {N}[G]$ that the subgraph $ H^{\prime }$ of H associated with G has m many disjoint rays.

The crucial point is that we have designed $\left \langle V_{i}^{\prime \prime },E_{i}^{^{\prime \prime }},f_{i}^{\prime \prime }\right \rangle _{i<m} $ so that $E^{\prime \prime }\cap N^{\left \langle \tau ,T\right \rangle }=\emptyset $ . Thus we may define a tree $T^{\prime }\subseteq T$ by simply removing all $\sigma $ such that $\sigma (n)=0$ for some $n\in E^{\prime \prime }$ . This change at most moves some edges $n\in U^{\left \langle \tau ,T\right \rangle }$ to $Y^{\left \langle \tau ,T^{\prime }\right \rangle }$ and makes $Y^{\left \langle \tau ,T^{\prime }\right \rangle }=Y=Y^{\left \langle \tau ,T\right \rangle }\cup E^{\prime \prime }$ as desired while keeping $N^{\left \langle \tau ,T^{\prime }\right \rangle }=N^{\left \langle \tau ,T\right \rangle }$ . If $\left \langle \tau ,T^{\prime }\right \rangle $ is a condition, it clearly extends $\left \langle \tau ,T\right \rangle $ and so we only have to verify that it is one. As we already have the required facts about $Y^{\left \langle \tau ,T^{\prime }\right \rangle }$ , $N^{\left \langle \tau ,T\right \rangle }$ , and $ U^{\left \langle \tau ,T^{\prime }\right \rangle } $ , we only need to supply witnesses $f^{\prime }$ and $A^{\prime }$ . As we have the witness f for $ \left \langle \tau ,T\right \rangle $ , to get $f^{\prime }$ it clearly suffices to compute for each v the number of $n>|\tau |$ such that $n\in E^{\prime \prime }-Y^{\left \langle \tau ,T\right \rangle }$ and $M(v,n)$ . We begin with the sequence $\left \langle V_{i}^{\prime \prime },E_{i}^{^{\prime \prime }},f_{i}^{\prime \prime }\right \rangle _{i<m}$ of disjoint rays such that $E^{\prime \prime }=\cup \{E_{i}^{^{\prime \prime }}|i<m\}$ . We first determine those i for which $v\in V_{i}^{\prime \prime }$ . (For the other $ j$ there are no $n\in E_{j}^{\prime \prime }$ with $M(v,n)$ .) As v appears in each of these $V_{i}^{\prime \prime }$ only once, there are at most two such edges in each of these $E_{i}^{\prime \prime }$ . Using this information we can determine for which edges $n\in E_{i}^{^{\prime \prime }}-Y^{\left \langle \tau ,T\right \rangle }$ we have $M(v,n).$ As the $ E_{i}^{\prime \prime }$ are disjoint, we can now simply add up the contributions (of one or two) from each $E_{i}^{\prime \prime }-Y^{\left \langle \tau ,T\right \rangle }$ with $v\in V_{i}^{\prime \prime }$ to get the amount we need to add to $f(v)$ to get the desired $f^{\prime }(v)$ . Finally, as $N^{\left \langle \tau ,T\right \rangle }=N^{\left \langle \tau ,T^{\prime }\right \rangle }$ and $U^{\left \langle \tau ,T^{\prime }\right \rangle }\subseteq U^{\left \langle \tau ,T\right \rangle }$ , we can take the witness A for $\left \langle \tau ,T\right \rangle $ to also be the desired witness $A^{\prime }$ for $\left \langle \tau ,T^{\prime }\right \rangle $ .

Proof Continue the iteration by the et-forcings that provide solutions to an instance of one of the LF $_{XY}$ principles through $\omega _{1}$ many steps in such a way that every H that appears as an instance at any $ \mathcal {N}_{\alpha }$ gets a solution in some $\mathcal {N}_{\beta +1}$ for $ \beta \geq \alpha $ . At limit levels we still act continuously: $\mathcal {N} _{\lambda }$ is the union of the $\mathcal {N}_{\alpha }$ for $\alpha <\lambda $ . Now if $H\in \mathcal {N}_{\omega _{1}}$ is an instance of the LF $ _{XY}$ principle then, not only does H appear in some $\mathcal {N} _{\alpha }$ , but so do all the witnesses for H containing m many disjoint rays for every m as there are only countably many $m\in \mathcal {N }$ and the full sequence is of length $\omega _{1}$ .

Proof Each of these four principles together with ACA $_{0}$ is a THA by Proposition 3.6. By Theorem 5.2, none of them imply ACA $ _{0}$ . Combining them and any other principles for which solutions can be added by et-forcings and even WKL by $\omega _{1}$ iterations is routine following the route indicated in Remark 4.14 and the comments proceeding it for $\omega $ length iterations.

Proof Take the solutions $Y_{i}$ given by $P^{\ast }$ and construct the sequence $ Y_{i,\sigma }$ for each finite (binary) string $\sigma $ with $Y_{i,\sigma }(n)=\sigma (n)$ for $n<|\sigma |$ and $Y_{i,\sigma }(n)=Y_{i}(n)$ for $ n\geq |\sigma |$ .

Notation 5.14. For a function f, finite string $\mu $ (or set X), and $\sigma $ a finite (binary) string we write $f_{\sigma }$ , $\mu _{\sigma }$ (or $X_{\sigma }$ ) to mean the function, finite string (or set) gotten by using $\sigma $ to define its initial segment of length $|\sigma |$ : $f_{\sigma }(i)=\sigma (i)$ for $i<|\sigma |$ , $f_{\sigma }(i)=f(i)$ for $i\geq |\sigma |$ and similarly for $\mu _{\sigma }$ and $X_{\sigma }$ . We write $f_{\sigma }^{[n]}$ for $ (f^{[n]})_{\sigma }$ and $X_{\sigma }^{[n]}$ for $(X^{[n]})_{\sigma }$ . Similarly, for a tree T we write $T_{\sigma }=\{\mu _{\sigma }|\mu \in T\}$ . We write $T_{\rho }^{\sigma }$ for $(T^{\sigma })_{\rho }$ where $ T^{\sigma }$ is defined in Notation 4.3.

Proof The implication $\Sigma _{1}^{1}$ -AC $\rightarrow \Sigma _{1}^{1}$ -AC $^{\kern1.2pt \ast } $ is immediate as solutions to instances of the former are also solutions to the same instance of the latter. The implications from a $^{\ast }$ version to the corresponding $^{-}$ one in RCA $_{0}$ are essentially instances of Proposition 5.4. Of course, the versions of TAC are simply special cases of the corresponding one for $\Sigma _{1}^{1}$ -AC. (The $ \left \langle T_{i}\right \rangle $ is absorbed into the set parameters.) For the equivalences in ACA $_{0}$ it thus suffices to show that TAC $ ^{-}\rightarrow \Sigma _{1}^{1}$ -AC $_{0}$ . First note that TAC follows from TAC $^{-}$ in ACA $_{0}$ as, given an instance of TAC and the f provided by TAC $^{-}$ , one can construct the one required by TAC recursively in the jump of f and the given sequence of trees. Now we have $\Sigma _{1}^{1}$ -AC by Proposition 5.13.

Finally that RCA $_{0}\vdash ~$ TAC $^{-}\rightarrow $ TAC $^{\ast }$ requires an argument that does not work for the $\Sigma _{1}^{1}$ -AC analogs. We are given a sequence $\left \langle T_{n}\right \rangle $ of trees such that $ \forall n\exists f(f\in \lbrack T_{n}])$ and an f such that $\forall n\exists m(f^{[m]}\in \lbrack T_{n}])$ and must produce a g such that $ \forall n\exists \sigma (g_{\sigma }^{[n]}\in \lbrack T_{n}]$ . To construct $ g^{[n]}$ we start by copying $f^{[0]}$ until we have an $s_{0}$ such that $ f^{[0]}\upharpoonright s_{0}+1\notin T_{n}.$ If we never find such an s we have computed a branch on $T_{n}$ and so $g^{[n]}=g_{\emptyset }^{[n]}$ is as required. If we find such an $s_{0}$ we switch to copying $f^{[1]}$ for inputs from $s_{0}$ onward. We continue until we once again fall off $T_{n}$ , i.e., $f^{[1]}\upharpoonright s_{1}+1\notin T_{n}$ . By the conclusion of TAC $^{-}$ from some point onward there is a fixed m such that we are copying $ f^{[m]}$ and $f^{[m]}\in \lbrack T_{n}]$ . We have thus constructed a g such that $g_{\sigma }\in \lbrack T_{n}]$ with $\sigma =f^{[m]}\upharpoonright s_{m-1}$ .

Proof If a given instance $\Phi (A,n,X)$ of $\Sigma _{1}^{1}$ -AC is invariant under finite changes then a solution for the same instance of $\Sigma _{1}^{1}$ -AC $^{\ast }$ is also one for $\Sigma _{1}^{1}$ -AC. In the other direction, given an instance $\Phi (A,n,X)$ of $\Sigma _{1}^{1}$ -AC $^{\ast }$ , consider the one $\Psi (A,n,X)\equiv \exists \sigma \Phi (A,n,X_{\sigma })$ for $\Sigma _{1}^{1}$ -AC. Clearly $\Psi (A,n,X)$ is closed under finite changes and any $\Sigma _{1}^{1}$ -AC solution for $\Psi $ is also a $\Sigma _{1}^{1}$ -AC $^{\ast }$ solution for the $\Sigma _{1}^{1}$ -AC $^{\ast }$ instance $\Phi (A,n,X)$ .

Proof By Proposition 5.16 it suffices to prove the theorem for $\Sigma _{1}^{1}$ -AC $^{\ast }$ . Given a countable model $\mathcal {N}$ of RCA $_{0}$ and an arithmetic $\Phi $ such that $\mathcal {N}\vDash \forall n\exists X\Phi (A,n,X)$ we define a forcing with conditions $\left \langle \tau ,T\right \rangle $ such that, in $\mathcal {N}$ , there is a finite set F and a sequence $\left \langle X_{i}|i\in F\right \rangle $ such that $\forall i\in F\Phi (A,i,X_{i})$ and for all $\sigma \in T$ , $|\sigma |>n\geq |\tau |$ with $n=\left \langle i,m\right \rangle $ for some $i\in F$ , $\sigma (n)=X_{i}(m)$ . (Otherwise there are no restrictions on $\sigma $ .)

It is now easy to see that the associated notion of forcing is et: If $ \left \langle \tau ,T\right \rangle $ is a condition and $\sigma \in T$ then $ \left \langle \sigma ,T^{\sigma }\right \rangle $ is a condition with the same F and $X_{i}$ which extends $\left \langle \tau ,T\right \rangle $ . It is also clear the sets $D_{i}$ of conditions such that i is a member of the associated F are dense. (Just thin out a given $\left \langle \tau ,T\right \rangle $ by choosing an $X_{i}$ such that $\Phi (A,i,X_{i})$ and keeping only those $\sigma \in T$ which satisfy the condition that $\sigma (\left \langle i,m\right \rangle )=X_{i}(m)$ for $\left \langle i,m\right \rangle \geq |\tau |$ .) Moreover, for any G on this thinned out tree, $G^{[i]}=^{\ast }X_{i}$ . Thus for any G generic for these $D_{i}$ , $ G $ is the desired witness for this instance of $\Sigma _{1}^{1}$ -AC $^{\ast } $ .