Constraining a dense sequence on a product space, one factor at a time

Question

Slogan: Given a sequence on $X\times Y$, can we choose subsequences to fix the limit in $X$ while leaving the behavior on $Y$ free?

Details: Suppose $X$ and $Y$ are topological spaces and $(x_n,y_n)_n$ is a sequence that is "limit-dense" in $X\times Y$, meaning that for every point $(x,y)\in X\times Y$, there is a subsequence $(x_m,y_m)_m$ that converges to $(x,y)$.

Does that imply the following: For every point $x\in X$, there is a subsequence $(x_m,y_m)_m$ such that $(x_m)_m$ converges to $x$ and $(y_m)_m$ is limit-dense in $Y$?

The answer is "yes" if $X$ is first-countable and $Y$ is second-countable. Those axioms give us the open rectangles that we need the subsequence to hit. But are those additional assumptions necessary?

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Edits:

Here's a proof of the earlier claim. Suppose $X$ is first-countable and $Y$ is second-countable. Let $U_1\supset U_2\supset\cdots\ni x$ be a descending countable local basis at $x$ and let $V_i$ be a countable basis on $Y$. Define a sequence of target rectangles $(T_n)_n$ to be $(U_1\times V_1, U_2\times V_1, U_2\times V_2, U_3\times V_1, U_3\times V_2, U_3\times V_3, \ldots)$. Now recursively define the subsequence $n_k$ by $n_0=0$ and for $k\geq 1$, $n_k$ is the first value greater than $n_{k-1}$ such that $(x_{n_k}, y_{n_k})\in T_k$. Then $x_{n_k}\to x$. And for any $y\in Y$, we can find a sequence of $V$s that are a descending countable local basis around $y$, and then a further subsequence of $y_{n_k}$ that stays within that basis.

For context, I'm working on an application where $X$ and $Y$ are really nice spaces, compact manifolds, so the claim is fairly obvious: For our target rectangles, we can cover $\{x\}\times Y$ with $1/n$-balls. I'm just curious how far we can take the principle for more general spaces.

Answer 1

With a diagonal construction, we can lower the bar to

$X$ and $Y$ first-countable

Let $x\in X$ be given and $U_0\supset U_1\supset U_2\supset\ldots \ni x$ a desceinding countable local basis at $x$. For each $m\in\Bbb N$, pick a subsequence $(x_{s_m(n)},y_{s_m(n)})$ with $(x_{s_m(n)},y_{s_m(n)})\to (x,y_m)$. Let $f(n)=n\bmod\lfloor\sqrt n\rfloor$ (or any other function $\Bbb N\to\Bbb N$ such that $\forall n\in\Bbb N\colon |f^{-1}(n)|=\infty$). As $\lim_{k\to\infty}s_{f(n)}(k)= \infty$ and $\lim_{k\to\infty}x_{s_{f(n)}(k)}= x$, we can define $s(n)$ recursively as $s(n)=s_{f(n)}(k)$, where $k=k(n)$ is minimal with $x_{s_{f(n)}(k)}\in U_n$ and $s_{f(n)}(k)>s(j)$ for all $j<n$. By the above choice, clearly $x_{s(n)}\to x$. Also, for each $m\in\Bbb N$, there is a subsequence of $\{y_{s(n)}\}_n$ that converges to $y_m$.

Now let $y\in Y$ and $V_0\supset V_1\supset V_2\supset\ldots \ni y$ a descending countable local basis at $y$. As there is a subsequence of the original sequence that converges to $(x,y)$, each $V_n$ contains some $y_m$, and as a subsequence of our constructed subsequence $\{y_{s(k)}\}_k$ converges to $y_m$, there are infinitely many $k$ with $y_{s(k)}\in V_n$. This allows us to pick a (sub-)subsequence $\{y_{s(t(n))}\}_n$ such that $y_{s(t(n))}\in V_n$ and consequently $y_{s(t(n))}\to y$.