Formula for slices of interior in a product space

Question

Let $X$, $Y$ be topological spaces, and for $x \in X$ let $\cdot_x$ denote the "slicing operation at $x$" which takes $S \subseteq X \times Y$ and returns the slice $S_x := \{ y\in Y : (x, y) \in S \} \subseteq Y$. Fix $x \in X$, and let $\mathcal{N}_x$ be a neighborhood basis at $x$. Prove that for any $S \subseteq X \times Y$: \begin{equation} (\mathrm{int} ~ S)_x = \bigcup_{N \in \mathcal{N}_x} \mathrm{int} \left ( \bigcap_{x' \in N} S_{x'} \right ). \end{equation}

Answer 1

OK, here's a possible solution, though possibly not exactly the solution you'd come up with just looking at the problem; and by "reading it backwards" you will probably sort of be able to see how I came up with the formula in the first place.

Fix $y \in Y$, and choose some neighborhood basis $\mathcal{N}_y'$ at $y$ (if nothing else, the entire set of neighborhoods of $y$ will suffice). Then $y$ is in the right hand side if and only if \begin{align} ~ & \exists N \in \mathcal{N}_x, ~ y \in \mathrm{int} \left( \bigcap_{x'\in N} S_{x'} \right) \\ \Leftrightarrow ~ & \exists N \in \mathcal{N}_x, ~ \exists N' \in \mathcal{N}_y', ~ N' \subseteq \bigcap_{x' \in N} S_{x'} \\ \Leftrightarrow ~ & \exists N \in \mathcal{N}_x, ~ \exists N' \in \mathcal{N}_y', ~ \forall y' \in N', ~ y' \in \bigcap_{x' \in N} S_{x'} \\ \Leftrightarrow ~ & \exists N \in \mathcal{N}_x, ~ \exists N' \in \mathcal{N}_y', ~ \forall y' \in N', ~ \forall x' \in N, ~ y' \in S_{x'} \\ \Leftrightarrow ~ & \exists N \in \mathcal{N}_x, ~ \exists N' \in \mathcal{N}_y', ~ \forall y' \in N', ~ \forall x' \in N, ~ (x', y') \in S \\ \Leftrightarrow ~ & \exists N \in \mathcal{N}_x, ~ \exists N' \in \mathcal{N}_y', ~ N \times N' \subseteq S. \end{align} Since $\{ N \times N' : N \in \mathcal{N}_x, ~ N' \in \mathcal{N}_y' \}$ forms a neighborhood basis for the product topology at $(x, y)$, this is further equivalent to $(x, y) \in \mathrm{int}~S$, which by definition is equivalent to $y$ being in the left hand side.