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The following definition is by Sinclair, G.E. A finitely additive generalization of the Fichtenholz–Lichtenstein theorem. Transactions of the American Mathematical Society. 1974;193:359-74.

A function $f : X \times Y \to R$ is double-limit continuous (DLC) if, whenever for a sequence $x_{m}$ in $X$ and a sequence $y_{n}$ in $Y$ both limits $$\lim_{m \to \infty}\lim_{n \to \infty}f(x_{m},y_{n})\text{ and }\lim_{n \to \infty}\lim_{m \to \infty}f(x_{m},y_{n})$$ exist, they are equal.

Consider the following special case: $X = \mathbb{N}$ is the set of natural numbers, $Y \subseteq \{0,1\}^{\mathbb{N}}$, and $f$ is defined by $f(x,y) = y_{x}$. Consider also the statements:

Statement 1: "There is an uncountable set $Y$ such that $f$ is DLC".

Statement 2: "There is a perfect set $Y$ such that $f$ is DLC".

Question: Is statement 1 false in ZFC? statement 2? Are the statements independent of ZFC?

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    $\begingroup$ So there is no requirement about when the double limits exist, or that they always both exist or both diverge, only that, when they both exist, they are equal? $\endgroup$ Commented Feb 20 at 13:31
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    $\begingroup$ @Lspice: exactly. IF both exist, THEN they are equal. $\endgroup$ Commented Feb 20 at 14:02
  • $\begingroup$ One tag of yours implicitly suggests the definition of a cardinal characteristic - the minimal cardinality $\kappa$ such that no $Y \in [2^\omega]^\kappa$ is DLC, let us call this number the double-limit discontinuity number $\mathfrak{d}\mspace{-1mu}\mathfrak{d}$. Bounding that from below by a run-of-the-mill cardinal characteristic would show that ZFC fails to refute Statement 1. $\endgroup$ Commented Aug 7 at 13:32

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ZFC refutes Statement 2.

Proof: We are going to use a variant of a Theorem of Blass.

Theorem(Blass): If $k, n < \omega$, $P$ is perfect but not empty and $\chi : [P]^k \longrightarrow n$ is a Borel(Baire also works) colouring of the $k$-tuples with $n$ colours, then there is a perfect nonempty $Q \subset P$ on which colour only depends on the splitting type of the $k$-tuple. In particular we can do with $(n - 1)!$ colours.

We call the perfect sets witnessing the truth of the theorem weakly homogeneous for $\chi$.

The reason we cannot do better is that in every perfect set there are unavoidable patterns of $k$-tuples. Say we colour triples, this is the case we need here anyway: If $x < y < z$ lexicographically we could have $|y - x| < |z - y|$ or $|z - y| < |y - x|$ or $|y - x| = |z - y|$. The last case is avoidable, every perfect set contains a perfect nonempty subset without a triple of that type but the first two have to appear in any perfect set.

We are going to be interested in a particular type of continuous triple colouring for the triples of the first type, i.e. the ones where the two leftmost points are closer to each other than the two rightmost ones. This means that the rightmost points splits of from the two leftmost at a lower level than where the two leftmost points split from each other. Let $s(x, y)$ be the longest common initial segment of the two leftmost points $x$ and $y$, i.e. their splitting node, and let $\ell(s)$ be its length. We are considering the colouring $\chi(\{x, y, z\}) = z(\ell(s(x,y)))$ which is continuous, even Lipschitz.

Now let $Q$ be a perfect weakly homogeneous set for $\chi$, let us say in colour $i$. We may pick $q \in Q$ so that if $\langle s_k | k < \omega\rangle$ is an enumeration of the splitting nodes which are initial segments of $q$, we have $s_{2k + j}\frown\langle j\rangle \sqsubset q$ for all natural numbers $k$ and both $j < 2$.

Let $\langle x_n | n < \omega\rangle$ be defined by $x_n = \ell(s_{2n + 1 - j})$ and let $\langle y_n | n < \omega\rangle$ be a sequence of elements of $Q$ such that $y_n \sqsupset s_{2n}\frown\langle 1\rangle$. This sequence converges to $q$ from the right without ever becoming $q$. Unless I am mistaken we have $$ 1 - i = \lim_{m \rightarrow \infty} \lim_{n \rightarrow \infty} f(x_m, y_n) \ne \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} f(x_m, y_n) = i. $$ Quod erat demonstrandum.

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  • $\begingroup$ I had to change the proof. In the previous version $q$ could just have ended up being the rightmost element of $Q$ in which case it would have been difficult to find a sequence approaching $q$ from the right. $\endgroup$ Commented Aug 8 at 8:24

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