Proof. Suppose f is Lipschitz. Then for each $x, y \in X^\omega $ and $k < \omega $ we have that $f(x) \upharpoonright k = f(y) \upharpoonright k$ if $x \upharpoonright k = y \upharpoonright k$ hence the map $\tilde {f}:X^{<\omega } \to X^{<\omega }$ defined by $\tilde {f}(t) = s$ if and only if there is an $x \in X^\omega $ with $t \lhd x$ and $s = f(x) \upharpoonright |t|$ is well defined. Moreover it is easily verified that it has the properties claimed.

Conversely if we have a function $\tilde {f}$ which is order and level preserving then define $f(x)$ to be $\bigcup _{k < \omega } \tilde {f}(x \upharpoonright k)$ . Again it’s easy to see that it has the advertised properties.

Proof. The point is that the homomorphism determined by f is in fact defined on the tree of finite sequences extending to elements of A, whose set of branches is simply the closure of A. More concretely, let $f:A \to X^\omega $ be Lipschitz. If $a \in \bar {A}$ then there is a sequence $\{a_n\}_{n < \omega } \subseteq A$ so that for all $n < \omega $ we have that $a \upharpoonright n = a_n \upharpoonright n$ . Fix one and define $\hat {f}(a)$ to be the unique b so that $b\upharpoonright n = f(a_n) \upharpoonright n$ . Note that the Lipschitz property implies that this is well defined since $a_m \upharpoonright n = a_n \upharpoonright n$ for $m> n$ and is moreover independent of the choice of $\{a_n\}_{n < \omega }$ since by Lemma 2.2 we only need the first n bits of $a_n$ to define the map. The last thing to check is that this is indeed a Lipschitz function on $\bar {A}$ . To see this, suppose $a, b \in \bar {A}$ and $k < \omega $ is such that $a\upharpoonright k = b\upharpoonright k$ . There are in A elements $a', b' \in A$ so that $a' \upharpoonright k = a\upharpoonright k$ and $b' \upharpoonright k = b\upharpoonright k$ . We have that $\hat {f} (a)\upharpoonright k =f(a')\upharpoonright k = f (b') \upharpoonright k = \hat {f} (b) \upharpoonright k$ as needed.

Proof. The map $\hat {f}$ is exactly as defined in the previous proof. Let us check that it is a bijection and both it and its inverse are also isometries. We start by checking that it is injective. Suppose that $a, b \in \bar {A}$ are distinct. Let $k < \omega $ be such that $a \upharpoonright k \neq b \upharpoonright k$ . By the way that $\hat {f}$ was defined this means that there are $a_k, b_k \in A$ so that $a_k \upharpoonright k = a\upharpoonright k$ and $b_k \upharpoonright k = b\upharpoonright k$ . Since f is an isometry we have that $\hat {f}(a) \upharpoonright k = f(a_k) \upharpoonright k \neq f(b_k) \upharpoonright k = \hat {f}(b) \upharpoonright k$ and hence $\hat {f}(a) \neq \hat {f}(b)$ .

For surjectivity suppose that $b \in \bar {B}$ and let $\{b_k\}_{k < \omega }\subseteq B$ be a sequence converging to b with $b\upharpoonright k = b_k \upharpoonright k$ . Since f is surjective there are $\{a_k\}_{k < \omega } \subseteq A$ so that $f(a_k) = b_k$ for all $k< \omega $ . Now if $j> k$ we have that $b_j \upharpoonright k = b_k \upharpoonright k$ and, since f is an isometry therefore $a_j \upharpoonright k = a_k\upharpoonright k$ and consequently $\{a_k\}_{k < \omega }$ is a Cauchy sequence. Moreover since X is discrete, the space $X^\omega $ is complete. Thus there is an $a = \lim _{k \to \infty } a_k$ . Now observe though that $\hat {f}(a) = b$ by the way the lift was defined.

Proving that both $\hat {f}$ and its inverse are isometries is a straightforward generalization of the observation that $\hat {f}$ is Lipschitz in the case of Lemma 2.3. Indeed fix $a, a' \in \bar {A}$ with sequences $\{a_k\}_{k < \omega }$ and $\{a_k'\}_{k < \omega }$ in A so that for all $j < \omega $ we have $a \upharpoonright j = a_j \upharpoonright j$ and $a' \upharpoonright j = a_j'\upharpoonright j$ . Then for all $k < \omega $ we have $a \upharpoonright k = a' \upharpoonright k$ if and only if $a_k' \upharpoonright k = a' \upharpoonright k$ if and only if $f(a_k) \upharpoonright k = f(a_k ') \upharpoonright k$ if and only if $\hat {f}(a) \upharpoonright k = \hat {f}(a') \upharpoonright k$ . The case of the inverse is identical.

Proof. This is a back and forth argument. Let $A, B \subseteq X^\omega $ be countableFootnote ⁴ and dense and enumerate them as $A = \{a_n\; | \; n < \omega \}$ and $B = \{b_n\; | \; n < \omega \}$ . We recursively define isometries $\{f_n\}_{n < \omega }$ so that the following hold for all $n < \omega $ :

1. (1) $\mathrm {dom}({f_n}) \supseteq \{a_0,\ldots , a_n\}$ and $\mathrm {range}(f_n) \supseteq \{b_0,\ldots , b_n\}$ .

2. (2) $f_n$ is an isometry in the sense that for each $a, a' \in \mathrm {dom}(f_n)$ and all $k < \omega $ we have that $a \upharpoonright k = a' \upharpoonright k$ if and only if $f(a) \upharpoonright k = f(a') \upharpoonright k$ .

Having accomplished this the desired isometry will be the lift of $\bigcup _{n < \omega } f_n$ obtained from Lemma 2.4. Thus we just need to carry out the recursive construction. The case of $n = 0$ is easy. We simply let $f_0 (a_0) = b_0$ , with this being the unique value in the domain. Given $f_n$ satisfying (1) and (2) above we need to find $f_{n+1} \supseteq f_n$ with $a_{n+1}$ in its range and $b_{n+1}$ in its domain. Without loss of generality let us assume that $a_{n+1} \notin \mathrm {dom}(f_n)$ and $b_{n+1} \notin \mathrm {range}(f_n)$ . If this is not true then we can ignore the “forth” or “back” subcases below, respectively. We begin with the “forth” case. Let $k < \omega $ be maximum so that there is an $a \in \mathrm {dom}(f_n)$ with $a_{n+1} \upharpoonright k = a \upharpoonright k$ —such a k exists since the domain of $f_n$ is finite. Let $i < \omega $ so that $a_i \in \mathrm {dom}(f_n)$ witnesses this k (such an i need not be unique—we will return to this point). Since B is dense, there is a $b \in B$ so that $b \upharpoonright k = f(a_i) \upharpoonright k$ . This is almost good enough but we need to choose b a little more carefully.

Here we return to the point that $a_i$ may not have been unique so that it agreed with $a_{n+1}$ for a maximal length. The claim states that we can find a single $b \in B$ which agrees with not only $f_n(a_i)$ for exactly the first k bits but indeed every such $a_j$ which was maximal in this sense.

Let us now fix a $b \in B$ as in the claim and check that $f_{n} \cup \{\langle a_{n+1}, b\rangle \}$ is an isometry. We need to show that for all $a_j \in \mathrm { dom}(f_n)$ and all k we have $a_j \upharpoonright k = a_{n+1} \upharpoonright k$ if and only if $f_n(a_j) \upharpoonright k = f_n(a_{n+1}) \upharpoonright k$ . There are two cases.

Case 1: $a_j$ is not maximally good. Then there is a minimum $l < k$ so that $a_j \upharpoonright l \neq a_{n+1} \upharpoonright l$ . We have that for all $m < l$ the equality $a_j \upharpoonright m = a_{n+1} \upharpoonright m = a_i \upharpoonright m$ holds (where $a_i$ is the maximally good element we fixed in the beginning of the proof). But then, by the fact that $f_n$ is an isometry we have $f_n (a_j) \upharpoonright m = f_n (a_i) \upharpoonright m = b\upharpoonright m$ by the choice of b. However $b \upharpoonright l = f_n(a_i) \upharpoonright l\neq f_n(a_j)\upharpoonright l$ again since $f_n$ is an isometry and the choice of b. Putting all this together we get that $a\upharpoonright m = a_j\upharpoonright m$ if and only if $b \upharpoonright m = f_n(a_j)\upharpoonright j$ , as needed.

Case 2: $a_j$ is maximally good. The case is somewhat similar but reduces to the choice of b.

The “back” part of the argument is entirely symmetric and left to the reader.

Proof. Fix a distinguished element $0_X \in X$ . For each $s \in X^{<\omega }$ let $O_s$ be a $\kappa $ sized set of elements of $[s]$ so that for all $k \geq |s|$ if $x \in O_s$ is such that $x(k) \neq 0_X$ then k is odd. Similarly let $E_s$ be the same with “odd” in the previous paragraph replaced by “even.”Footnote ⁵ Note that for each $s \in X^{<\omega }$ if $x, y \in O_s$ then the first place that they differ is even. Similarly if $x, y \in E_s$ then the first place they differ is odd. Let $O = \bigcup _{s \in X^{<\omega }} O_s$ and $E = \bigcup _{s \in X^{<\omega }} E_s$ . Clearly these are $\kappa $ -dense. Suppose that $A \subseteq O$ is uncountable (the case of $A \subseteq E$ is symmetric). If $f:A \to E$ is any function at all there is some $s \in \omega ^{<\omega }$ by pigeonhole so that $f^{-1}[E_s]$ is uncountable. Applying pigeonhole again, there is a $t \in \omega ^{<\omega }$ so that $f^{-1}[E_s] \cap O_t$ is uncountable and in particular has size greater than $2$ . But if $x, y \in O_t$ then the first k at which they differ is odd while if $f(x), f(y) \in E_s$ then the first place they differ is even hence f is not an isometry.

Proof. Fix $f:2^\omega \to 2^\omega $ , Lipschitz. We begin with $(1)$ implies $(2)$ . Suppose f is surjective and, towards a contradiction assume there are $x, y \in 2^\omega $ and $k \in \omega $ so that $x(k) \neq y(k)$ but $f(x) \upharpoonright k+1 = f(y) \upharpoonright k+1 := t$ . In other words x and y witness that f is not an isometry as f maps them properly closer together. In this case then the map $\tilde {f}$ from Lemma 2.2 maps both $x \upharpoonright k+1$ and $y \upharpoonright k+1$ (which are assumed to be distinct) onto this same t. This implies however that $\tilde {f} \upharpoonright 2^{k+1}$ is not injective and hence not surjective (since the set is finite) and therefore $\mathrm {range}(f)$ is disjoint from some basic open $[s]$ with $s \in 2^{k+1}$ and in particular is not surjective.

Now we prove $\neg (1)$ implies $\neg (2)$ . Since f is continuous, the image of f is compact and hence closed in $2^\omega $ . As such, if f is not surjective, then there is some $k < \omega $ and some finite sequence $s \in 2^k$ so that $\mathrm {range} (f) \cap [s] = \emptyset $ . It follows then that the map $\tilde {f} \upharpoonright 2^k$ is not a surjection and hence there are distinct sequences $t_0, t_1 \in 2^k$ so that $\tilde {f}(t_0) = \tilde {f}(t_1)$ . But then if $t_0 \lhd x_0$ and $t_1 \lhd x_1$ we have that $d(x_0, x_1) \geq \frac {1}{2^k}$ while $d(f(x_0), f(x_1)) < \frac {1}{2^k}$ so f is not an isometry.

Proof. Assume $\mathsf {BA}_{\mathrm {Lip}}(\omega ^\omega )$ and let $A, B \subseteq 2^\omega $ be $\aleph _1$ -dense. Let $A' \subseteq \omega ^\omega $ be $\aleph _1$ -dense extending A (which is possible since A is subset of $\omega ^\omega $ trivially since it is a subset of $2^\omega $ ). We want to apply $\mathsf {BA}_{\mathrm { Lip}}(\omega ^\omega )$ to $A'$ and a $B'$ we will define below in terms of B. Towards this we need the following.

Now let $f:A' \to B'$ be an injective Lipschitz function—the existence of such is guaranteed by $\mathsf {BA}_{\mathrm {Lip}}(\omega ^\omega )$ . Note that $\pi :\omega ^\omega \to 2^\omega $ is also Lipschitz in the sense that if $x, y \in \omega ^\omega $ and $k < \omega $ so that $x \upharpoonright k = y \upharpoonright k$ then $\pi (x) \upharpoonright k = \pi (y) \upharpoonright k$ . Consider the restriction $\pi \circ f \upharpoonright A: A \to \pi (B')$ . Since $\pi (B') = B$ the range of this function is B. This is a Lipschitz function from A to B which is moreover injective since $\pi $ is a bijection on $B'$ . Thus this is the required map to witness $\mathsf {BA}_{\mathrm {Lip}}(2^\omega )$ .

Proof. Assume $\overline {\mathsf {BA}}_{\mathrm {Lip}}(\omega ^\omega )$ . As in the proof of Lemma 3.5 let $A, B \subseteq 2^\omega $ . Let $B'$ be defined exactly as in the proof of Lemma 3.5—i.e., $B' \subseteq \omega ^\omega $ is $\aleph _1$ -dense and $\pi :B' \to B$ is a bijection. For each $s \in {\omega }^{<\omega }$ let $A_s = \{s^\frown a\; | \; a \in A\}$ . Clearly $A' = \bigcup _{s \in \omega ^{<\omega }} A_s$ is $\aleph _1$ -dense. Applying $\overline {\mathsf {BA}}_{\mathrm {Lip}}(\omega ^\omega )$ , fix $f:A' \to B'$ a Lipschitz bijection. For each $s \in \omega ^{<\omega }$ let $g_s:A \to A_s$ be the obvious map, i.e., $g_s (a) = s^\frown a$ . Note that for each $s \in \omega ^{<\omega }$ the map $g_s$ is Lipschitz since if $a \upharpoonright k = b \upharpoonright k$ then $g_s (a) \upharpoonright k + |s| = b \upharpoonright k + |s|$ . Finally let $f_s:A \to B$ be $\pi \circ f \circ g_s$ . Again this is a composition of Lipschitz functions hence Lipschitz. To complete the proof we just have to see that $\bigcup _{s \in \omega ^{<\omega }} f_s"A$ covers B. Let $b \in B$ . Then $b' = \pi ^{-1}(b)$ is equal to $f(a')$ for some $a' \in A'$ since f was assumed to be surjective. But now $a' = s^\frown a$ for some s and $a \in A$ by definition and therefore $f_s (a) = b$ as needed.

Fact 3.9. Assume $\mathsf {BA}$ .

1. (1) (Abraham–Rubin–Shelah [Reference Abraham, Rubin and Shelah1, Theorem 7.1]) $2^{\aleph _0} = 2^{\aleph _1}$ .

2. (2) (Todorcevic, see [Reference Todorčević18]) $\mathfrak b> \aleph _1$

Fact 3.11 (Baumgartner, see page 414 of [Reference Baumgartner7])

If $2^{\aleph _0} < \aleph _{\omega _1}$ then there is an almost disjoint family on $\omega _1$ of size $2^{\aleph _1}$ .

Proof of Lemma 3.10

Fix an $\aleph _1$ -dense $A \subseteq 2^\omega $ and fix an almost disjoint family on $\omega _1$ of size $2^{\aleph _1}$ , say $\mathcal X$ . The existence of such is the only application of the assumption $2^{\aleph _0} < \aleph _{\omega _1}$ . Suppose towards a contradiction that moreover $2^{\aleph _0} < 2^{\aleph _1}$ . Partition A into $\omega _1$ many countable dense subsets $\{A_\alpha \}_{\alpha \in \omega _1}$ , i.e., for $\alpha \neq \beta $ then $A_\alpha \cap A_\beta = \emptyset $ and $A = \bigcup _{\alpha \in \omega _1} A_\alpha $ (the existence of such a partition is possible by $\aleph _1$ -density). For each $X \in \mathcal X$ let $A_X = \bigcup _{\alpha \in X} A_\alpha $ . This is also $\aleph _1$ -dense. Note that since $X, Y \in \mathcal X$ are almost disjoint we have that $A_X \cap A_Y$ have countable intersection as well.

By $\mathsf {BA}_{\mathrm {Lip}}(2^\omega )$ , for each $X \in \mathcal X$ there is a Lipschitz function $f_X:2^\omega \to 2^\omega $ so that $f_X \upharpoonright A$ injects into $A_X$ . Since there are only continuum many continuous functions on $2^\omega $ there are distinct $X, Y \in \mathcal X$ for which $f_X = f_Y$ . But this is a contradiction since the image of A under $f_X = f_Y$ must have size $\aleph _1$ but gets mapped to two sets whose intersection is countable.

Fact 3.13 (Theorem 1 of [Reference Bartoszynski and Shelah5])

There is a set $X \subseteq 2^\omega $ of size $\mathfrak b$ so that no continuous $f:X \to \omega ^\omega $ can map X onto an unbounded set.

Proof of Lemma 3.12

Assume $\mathsf {BA}_{\mathrm {Lip}}(2^\omega )$ and, towards a contradiction assume $\mathfrak b = \aleph _1$ . Let $B = \{b_\alpha \; | \; \alpha \in \omega _1\}$ be a $\mathfrak b$ -scale, i.e., a set which is unbounded and so that if $\alpha < \beta $ then $b_\alpha \leq ^* b_\beta $ . The assumption on $\mathfrak b$ guarantees the existence of such a set. Moreover, by making finite modifications to the $b_\alpha $ ’s (which does not affect the fact that we have a scale), we can also assume without loss of generality that B is $\aleph _1$ -dense in $\omega ^\omega $ . Since $2^\omega $ can be written as $\omega ^\omega \cup C$ with C a (disjoint) countable dense set, we can treat $B \subseteq 2^\omega $ and in fact assume B is $\aleph _1$ -dense in $2^\omega $ . Let $X \subseteq 2^\omega $ be a set of size $\aleph _1$ as in Fact 3.13 and let $A \supseteq X$ be $\aleph _1$ -dense.Footnote ⁶ Let $f:X \to B$ be a continuous, injective function. Of course our assumption of $\mathsf {BA}_{\mathrm {Lip}}(2^\omega )$ gives that f is in fact Lipschitz but this is not even needed. We now immediately have a contradiction however to the defining property of X since any $\aleph _1$ -sized subset of B, in particular the image of X under f, will be again unbounded.

Proof. Assume $\mathsf {CH}$ . Fix A and B which are $\aleph _1$ -dense in $\omega ^\omega $ and let $\mathbb P'$ be the set of finite partial Lipschitz injections $f:A \to B$ . In other words $p \in \mathbb P$ is a finite partial injection from A to B so that for all $a, a' \in \mathrm {dom}(p)$ if $a \upharpoonright k = a' \upharpoonright k$ then $p(a) \upharpoonright k = p(a') \upharpoonright k$ for all $k < \omega $ . We will find a ccc suborder of $\mathbb P'$ . First, let S be the set of all x which are a finite set of pairs $(s_i, t_i) \in (\omega ^{<\omega })^2$ so that every $(s_i, t_i) \in x$ is in $(\omega ^k)^2$ for some $k < \omega $ and x is separating, i.e., if $(s, t), (u, v) \in x$ then s and u are incomparable and similarly for t and v. Equivalently $[s] \cap [u] = \emptyset $ and the same for v and t. Note that every $x \in S$ is essentially a basic open around some $p \in \mathbb P'$ (viewed as an element of $((\omega ^\omega )^2)^{|p|}$ ) which is refined enough that no two elements in the domain (resp. the range) are in the same ball. Let $\mathbb P'(x)$ be the set of all $p \in \mathbb P'$ so that $|p| = |x|$ and for each $a \in \mathrm {dom}(p)$ there is a unique $(s, t) \in x$ with $(a, p(a)) \in [s] \times [t]$ . Of course there are many x so that $p \in \mathbb P'(x)$ for each $p \in \mathbb P'$ .

Using $\mathsf {CH}$ , enumerate all finite or countable subsets of $\mathbb P '$ which are contained in some $\mathbb P ' (x)$ as $\{d_\alpha \; | \; \alpha < \omega _1\}$ . For each such, let $x_\alpha $ be such that $d_\alpha \subseteq \mathbb P'(x_\alpha )$ . Let $c_\alpha $ be the set of all $p \in \mathbb P'$ so that for every $x \in S$ if $p \in \mathbb P'(x)$ then there is $q \in d_\alpha \cap \mathbb P'(x)$ . Note that $c_\alpha $ is in essence the closure of $d_\alpha $ in $\mathbb P '$ in the sense that it is literally the topological closure of $d_\alpha $ in $(\omega ^\omega )^{2|x|}$ intersected with the $|x|$ -length conditions in $\mathbb P '$ .

We note the interesting case of this is when U is uncountable.

We can now construct the partial order $\mathbb P$ . We will find partitions $A = \bigcup _{\alpha < \omega _1} A_\alpha $ and $B = \bigcup _{\alpha < \omega _1} B_\alpha $ so that $\mathbb P = \{p \in \mathbb P' \; | \; \forall \alpha \, p\upharpoonright A_\alpha :A_\alpha \to B_\alpha \}$ is ccc and each $A_\alpha $ and $B_\alpha $ are countable and dense. In fact, we don’t need to assume anything about the B partitionFootnote ⁷ so let $B = \bigcup _{\alpha < \omega _1} B_\alpha $ be a partition of B into countable dense sets and for each $\alpha < \omega _1$ enumerate $B_\alpha = \{b^\alpha _i\; | \; i < \omega \}$ . Suppose that $\beta < \omega _1$ and for all $\alpha < \beta $ we have already defined $A_\alpha $ . We work on defining $A_\beta $ .

We need to define countable, dense $\{a^\beta _n\; | \; n < \omega \} \subseteq A \setminus \bigcup _{\alpha < \beta } A_\alpha $ . Let $a^\beta _0$ be the least element of A not in any $A_\alpha $ so far (relative to some fixed enumeration in order type $\omega _1$ ). We do this simply to ensure the construction covers A. Let $\{s_n\; | \; n < \omega \}$ enumerate $\omega ^{<\omega }$ so that $a^\beta _0 \in [s_0]$ and for every $i < \omega $ we have $b^\beta _i \in [s_i]$ (we can choose any enumeration we like so we can do this—at worst we can let $s_0 = [\emptyset ]$ ). Suppose now that $\{a^\beta _i\; | \; i<n + 1\}$ have been defined and $a_i^\beta \in [s_i]$ . We define $a^\beta _n$ as follows. For each p in the already constructed part of $\mathbb P$ , i.e., the set $\{p \in \mathbb P\mid \mathrm {dom}(p) \subseteq \bigcup _{\alpha < \beta } A_\alpha \cup \{a_i^\beta \mid i < n+1\}\}$ , and each $i < n+1$ , $\alpha < \beta $ , $(s, t) \in x_\beta $ , let $X_{i, p, \beta , (s, t)}$ be the following set:

$$ \begin{align*}\{a \in A \cap [s]\; | \; p \cup (a, b^\beta_i) \in c_\beta \, \land b^\beta_i \in [t]\}\end{align*} $$

if this set is countable and let it be the empty set otherwise. Now choose $a^\beta _n$ to be any element of $[s_n]$ in A not already constructed nor in any of the $X_{i, p, \beta , (s, t)}$ ’s. Note that by $\aleph _1$ -density this is possible.

Let $\mathbb P$ be the set of $p \in \mathbb P'$ so that for all $\alpha < \omega _1$ we have $p\upharpoonright A_\alpha $ maps into $B_\alpha $ . Now we need to check that the construction works—i.e., that $\mathbb P$ is ccc. First we make a slight technical reduction. Note that $\mathbb P$ has a dense subset consisting of elements p so that if $\beta $ is greatest with $p \cap A_\beta \times B_\beta $ non-empty then there is a unique $a \in \mathrm {dom}(p) \cap A_\beta $ and if $a = a^\beta _k$ and $p(a) = b^\beta _l$ then $l < k$ . We call this unique element of $A_\beta \times B_\beta $ the last constructed element of p. We work with this dense subset. Towards a contradiction assume that $U \subseteq \mathbb P$ is an uncountable antichain. By standard arguments we can assume that there is an $n < \omega $ so that $|p| = n$ for all $p \in U$ , this n is minimal for which there is an antichain like this. Also, for each $p \neq q \in U$ we can assume $\mathrm {dom}(p) \cap \mathrm {dom}(q) = \mathrm {range}(p) \cap \mathrm {range}(q) = \emptyset $ . To see this, first apply a $\Delta $ -system argument to the sets of the form $\{\alpha \mid \mathrm {dom}(p) \cap A_\alpha \neq \emptyset \}$ for $p \in U$ . Note that, by thinning out to such a $\Delta $ -system, we ensure that there is a finite set of $\alpha $ for which there could be an overlap in either the domain or the range of two distinct $p, q \in U$ . Now, applying the pigeon hole principle plus the fact that there are only countably many possible targets for any given $a \in A$ by the way we constructed the poset, we can thin out further and obtain an uncountable antichain where the elements pointwise agree on their (shared and unique) common domain. However now if two such conditions are not compatible then they remain so with this shared piece removed—which would contradict the minimality of n were it not empty. Moreover, we can assume that there is an $\alpha < \omega _1$ so that $d_\alpha \subseteq U \subseteq \mathbb P'(x_\alpha )$ and there are $(s, t) \in x_\alpha $ so that every $p \in U$ has its last constructed pair in $[s]\times [t]$ (with respect to the construction given above). Moreover in that last constructed pair we know, because of the dense subset we restricted to, that the domain element is the last constructed one. Let $Q = \{p \setminus [s]\times [t] \; | \; p \in U\}$ and let $B' = B \setminus \bigcup _{p \in d_\alpha } \mathrm {range}(p)$ .

Case 1: $n = 1$ . In this case we claim that for every $b \in B'$ the set $Z_b$ of all a so that $\{(a, b)\} \in c_\alpha $ is a singleton. Indeed, suppose $a, a' \in Z_b$ and let k be least with $a (k) \neq a'(k)$ . Then there are $(a_0, b_0)$ and $(a_0', b_0')$ in $d_{\alpha }$ with $b_0 \upharpoonright k + 1 = b_0' \upharpoonright k+1 = b \upharpoonright k+1$ and $a_0 \upharpoonright k + 1 = a \upharpoonright k+1$ and $a_0'\upharpoonright k+1 = a' \upharpoonright k+1$ . But then these two conditions are compatible contradicting the assumption on U.

Case 2: $n> 1$ . We need some notation—given a condition $p \in \mathbb P$ and a natural number $l < \omega $ let $p \upharpoonright l$ denote the set of $\{(a \upharpoonright l, p(a) \upharpoonright l)\; | \; a \in \mathrm {dom}(p)\}$ . We claim that for all but countably many $q \in Q$ and every $b \in B'$ the set $Z_b$ of a so that $q \cup (a, b) \in c_\alpha $ is countable. Suppose not and let $Q'$ be an uncountable subset of Q with the property above. By the minimality of n we have to be able to find compatible $q_0, q_1 \in Q'$ . Let l be large enough so that every element in the ranges of $q_0$ and $q_1$ (and hence also their domains by the Lipschitz property) are separated. Note that now any two $q', r'$ so that $q' \upharpoonright l = q_0 \upharpoonright l$ and same for $q_1$ and $r'$ will be compatible. Now let $a_0$ and $a_1$ be so that $q_0 \cup (a_0, b) \in c_\alpha $ and the same for $q_1 \cup (a_1, b)$ and without loss of generality we may assume that $a_0 \upharpoonright l \neq a_1 \upharpoonright l$ and indeed l separates all the points in $q_i \cup (a_i, b)$ . Now let $r_0, r_1 \in d_\alpha $ so that $r_0 \upharpoonright l = q_0 \cup (a_0, b) \upharpoonright l$ and the same for $r_1$ with $q_1$ and $a_1$ . Then again $r_1$ and $r_0$ must be compatible.

Now it follows from the above two cases that either way we must find $p \in U$ so that the $(a, b) \in [s]\times [t] \cap p$ are in some $A_\beta \times B_\beta $ with $\beta $ above $\alpha $ and moreover for every $b \in B'$ the set $Z_b$ defined above with respect to $p \setminus [s] \times [t]$ is countable. But then a could not be chosen as it was by the construction of the partition. This completes the proof of the ccc.

To finish the proof of Lemma 4.1 we need to check that $\mathbb P$ forces the existence of a Lipschitz bijection from A to B. Let $G \subseteq \mathbb P$ be generic and work in $V[G]$ . Clearly $g:= \bigcup G$ is a Lipschitz injection from some subset of A into B. We need to show that in fact the domain is all of A and the range is all of B. Thus it remains to show that for all $a \in A$ and $b \in B$ there is a dense set of conditions q so that $a \in \mathrm {dom}(q)$ and $b \in \mathrm {range}(q)$ . Let $a \in A$ , $b \in B$ , and $p \in \mathbb P$ . Without loss of generality assume that $a \notin \mathrm {dom} (p)$ and $b \notin \mathrm {range}(p)$ . Let $\alpha , \beta < \omega _1$ so that $a \in A_\alpha $ and $b \in B_\beta $ . By density plus the finitenessFootnote ⁸ of p there are $k < \omega $ so that for no $a' \in \mathrm {dom}(p)$ do we have $a' (0) = k_0$ and there is a $c \in A_\beta $ so that $c(0) = k$ . Set $p' = p \cup \{\langle c, b\rangle \}$ . Note that this is still Lipschitz since c differs from every element of the domain of p on the first coordinate. Let $l < \omega $ now be maximal so that there is an $a' \in \mathrm {dom}(p ')$ with $a \upharpoonright l = a' \upharpoonright l$ . Let $j < \omega $ so that no $b' \in \mathrm {range}(p')$ is such that $b' (l) = j$ . Again by density there is a $d \in B_\alpha $ so that $d \upharpoonright l+1 = [p'(a') \upharpoonright l]^\frown j$ . Let $q = p' \cup \{\langle a, d\rangle \}$ . An easy verification shows that this is still Lipschitz and hence a condition.

Fact 6.4. $P_T$ is ccc if and only if T has no cofinal branch.

Fact 6.7. If $\kappa $ is an uncountable cardinal and $\mathbb P$ is the standard forcing to add either $\kappa $ many random reals or $\kappa $ many Cohens then any partition of the generic reals into two pieces A and B will Lipschitz avoid one another.

Proof. Fix $A, B \subseteq \omega ^\omega $ so that A Lipschitz avoids B and let $\langle \mathbb P_\alpha , \dot {\mathbb Q}_\alpha \; | \; \alpha \in \delta \rangle $ is a finite support iteration of ccc forcing notions so that for each $\alpha < \delta $ we have that $\Vdash _\alpha $ “ $\dot {\mathbb Q}_\alpha $ preserves that $\check {A}$ Lipschitz avoids $\check {B}$ .” We need to show that $\mathbb P_\delta $ forces that A Lipschitz avoids B. The proof is by induction on $\delta $ . The successor case is by assumption so we check the limit case. Assume towards a contradiction that $\Vdash _\delta $ “ $\dot {Z} \subseteq \check {A}$ is uncountable and $\dot {f}:\dot {Z} \to \check {B}$ is Lipschitz with uncountable range. There are two subcases.

Case 1: $\delta $ has countable cofinality. Let $\delta _n \nearrow \delta $ be a strictly increasing sequence of ordinals with $n \in \omega $ . Let $G_\delta $ be $\mathbb P_\delta $ -generic and for each $n < \omega $ let $G_n = G_\delta \cap \mathbb P_{\delta _n}$ be the $\mathbb P_{\delta _n}$ -generic given by the subforcing. Work momentarily in $V[G_\delta ]$ . Let $f_n$ be the partial function from X to Y given by $f_n(x) = y$ if and only if there is a condition $p \in G_n$ forcing that $\dot {f}(\check {x}) = \check {y}$ . Note that by the finite support we have that

$$ \begin{align*}\Vdash_{\delta} \dot{f}^{G_\delta} =\bigcup_{n < \omega} f_n.\end{align*} $$

Moreover observe that $f_n \in V[G_n]$ , where, by inductive assumption, A still Lipschitz avoids B. By the pigeonhole principle, there is an $n<\omega $ so that $f_n$ has uncountable domain. But being Lipschitz is absolute so this $f_n$ is already a counterexample to avoiding, which is a contradiction.

Case 2: $\delta $ has uncountable cofinality. In $V[G_\delta ]$ by Lemma 2.3 we can lift $\dot {f}^{G_\delta }$ to a function $\hat {f}$ whose domain is $\bar {Z}$ . Since $\bar {Z}$ is closed, both $\hat {f}$ and $\bar {Z}$ are coded by reals hence by the ccc plus finite support there is a $\gamma < \delta $ so that $\bar {Z}, \hat {f} \in V[G_\gamma ]$ . Work in $V[G_\gamma ]$ for such a $\gamma $ and note by assumption that A still Lipschitz avoids B in $V[G_\gamma ]$ . Observe first that if there are $x \in A$ and $y \in B$ so that some condition $p \in \mathbb P_{\gamma , \delta }$ forces that $\dot {f}(\check {x}) = \check {y}$ then by Schoenfield absoluteness in $V[G_\gamma ]$ we already have that $\hat {f}(x) = y$ . Moreover in $V[G_\delta ]$ we have that $\hat {f}$ is Lipschitz—but this is also absolute since to be Lipschitz is a $\Pi ^1_2$ statement with the parameters $\hat {f}$ and $\bar {Z}$ : $\forall x, y \in \bar {Z} \, \forall k \in \omega \; ( x \upharpoonright k = y \upharpoonright k \to \hat {f}(x) \upharpoonright k = \hat {f} (x) \upharpoonright k)$ . Therefore in $V[G_\gamma ]$ we have that $\hat {f}$ is Lipschitz and its domain includes uncountably many elements of A which are getting mapped into B. Therefore $\hat {f}\upharpoonright (A \cap \hat {f}^{-1}(B))$ is a Lipschitz map with uncountable range from a subset of A into B, contradicting avoiding.

Proof. Suppose towards a contradiction that $p \in \mathbb P$ forces that $\dot {Z} \subseteq A$ has size $\omega _1$ and there is a $\dot {f}:\dot {Z} \to B$ which is injective and Lipschitz. Concretely let $\{\dot {z}_\alpha \; | \; \alpha < \omega _1\}$ be names for the elements in some enumeration. For each $\alpha < \omega _1$ let $p_\alpha \leq p$ be a condition so that there is some $a_\alpha \in A$ and $b_\alpha \in B$ with

$$ \begin{align*}p_\alpha \Vdash \check{a}_\alpha = \dot{z}_\alpha \; \land \dot{f}(\check{a}_\alpha) = \check{b}_\alpha.\end{align*} $$

By the Knaster property there is an uncountable $K \subseteq \omega _1$ so that the set of all $\{p_\alpha \; | \; \alpha \in K\}$ are pairwise compatible.Footnote ¹⁰ Let $g_K:\{a_\alpha \; | \; \alpha \in K\} \to \{b_\alpha \; | \; \alpha \in K\}$ be defined by $g_K (a_\alpha ) = b_\alpha $ for all $\alpha \in K$ . By compatibility of the conditions this is a Lipschitz injection from an uncountable subset of A into B contradicting avoiding.

Proof. Fix T as in the statement of the lemma. Suppose $\dot {f}$ is a $\mathbb P_T$ -name for a Lipschitz function from an uncountable subset of A in B with uncountable range. Like in the proof of Lemma 6.8, Case 2, in the generic extension we can lift $\dot {f}$ to a $\hat {f}$ which is Lipschitz and has closed domain. By Theorem 6.5, there is a $\sigma $ -centered subforcing of $\mathbb P_T$ , say $\mathbb Q$ which adds the closed set and the Lipschitz function on it. However in this subextension we have that A Lipschitz avoids B since $\sigma $ -centered forcing notions are in particular Knaster and we can therefore apply Lemma 6.9. We are now though in exactly the same situation as in Case 2 of Lemma 6.8 and the same contradiction occurs.

Proof of Theorem 6.1

Start in a model of $\mathsf {CH}$ and fix A and B which are $\aleph _1$ -dense sets of $2^\omega $ so that A Lipschitz avoids B. By Lemmas 6.8–6.10 it follows that the standard iteration for forcing $\mathsf {MA} (\mathrm {Knaster}) +$ “every Aronszajn tree is special” will force these facts and A will still Lipschitz avoid B hence completing the proof.

Question 1. Are there further implications between $\mathsf {BA}_{\mathrm {Lip}} (\omega ^\omega )$ , $\mathsf {BA}_{\mathrm {Lip}} (2^\omega )$ , $\overline {\mathsf {BA}}_{\mathrm {Lip}} (\omega ^\omega )$ , and $\overline {\mathsf {BA}}_{\mathrm {Lip}} (2^\omega )$ than those given in Theorem 3.4? In particular are they all equivalent? Do the weak and strong versions equate? Do the Baire and Cantor versions equate?

Question 2. Do either $\mathsf {BA}_{\mathrm {Lip}} (\omega ^\omega )$ or $\mathsf {BA}_{\mathrm {Lip}} (2^\omega )$ imply $2^{\aleph _0} = 2^{\aleph _1}$ ? Do either of $\mathsf {BA}_{\mathrm {Lip}} (\omega ^\omega )$ or $\mathsf {BA}_{\mathrm {Lip}} (2^\omega )$ imply $\mathrm {add}(\mathcal N)> \aleph _1$ ?

Question 3. Do any of the Lipschitz axioms imply $\mathfrak {p}> \aleph _1$ ?

Question 4. Does $\mathsf {MA}$ imply $\mathsf {BA}_{\mathrm {Lip}}(2^\omega )$ ?

Question 5. Are there provable relations between $\mathsf {BA}$ and the Lipschitz variations?