Suppose $X$ and $Y$ are topological spaces and $U \subseteq X \times Y$.
Now, I managed to prove that whenever $U$ is open all "slices" along $Y$ are also open:

\begin{equation} U \; \text{is open} \implies \forall \, x \in X : \left \{ y \in Y : \left( x, y \right) \in U \right \} \; \text{is open} \end{equation}

Obviously the same applies to slices along $X$.

Does the converse hold? That is, suppose that for a given set $U \subseteq X \times Y$ all slices along $X$ and $Y$ are open, can we conclude that $U$ is open?
My gut feeling tells me, that we need something like Hausdorffness to ensure that we have sufficiently "small" open sets, but I fail to prove it (or disprove it for that matter).

My idea so far is quite simple. If $U$ is empty then all is well, so suppose there is an element $\left( x, y \right) \in U$.
We can then look at the neighbourhood filters $\mathcal{F}_x$ and $\mathcal{F}_y$ of $x$ and $y$ respectively. Then the statement is equivalent to the existence of some $A \in \mathcal{F}_x$, $B \in \mathcal{F}_y$ such that

\begin{equation} A \times B \subseteq U \end{equation}

This seems to me impossible to prove without further assumptions... What assumptions are necessary and sufficient? Uniform topology comes to mind as it assures some symmetry between openness in $X$ and $Y$ directions.

  • $\begingroup$ @Eric: Whoops, right. $\endgroup$ May 26 '17 at 20:23
  • $\begingroup$ Consider the complement in $\mathbb{R}^2$ of $\{(1,1), (1/2, 1/2), (1/3, 1/3), \ldots \}$ $\endgroup$ May 26 '17 at 21:26
  • $\begingroup$ Can you please share how you prove $U$ is open all "slices" along $Y$ are also open? $\endgroup$ May 27 '19 at 5:48
  • $\begingroup$ By the product topology, $(x, y) \in U \implies \exists \mathrm{open} A \in X, B \in Y : (x, y) \in A \times B \subseteq U$. Thus a slice along $Y$ that contains $y$ has to contain $B$. Hence to every $y$ in a slice, there is an open neighbourhood also contained in the slice. $\endgroup$
    – iolo
    May 29 '19 at 7:28

This is almost always false. To put this in a perhaps more familiar context, note that the collection of all sets $U\subseteq X\times Y$ whose slices are open is a topology; call it $T$. Now note that a map $f:X\times Y\to Z$ is continuous with respect to $T$ iff it is continuous on each coordinate separately: that is, iff for each $x\in X$, $y\mapsto f(x,y)$ is continuous, and for each $y\in Y$, $x\mapsto f(x,y)$ is continuous. So your question is equivalent to asking whether every separately continuous function on a product is actually jointly continuous (i.e., continuous with respect to the product topology).

As multivariable calculus students learn, this is false even in the most familiar contexts. For instance, taking $X=Y=Z=\mathbb{R}$, the function $f(x,y)=\frac{xy}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$ is continuous in each variable separately but is not continuous jointly at $(0,0)$. To get a direct counterexample to your question, you can take $U=f^{-1}(\mathbb{R}\setminus\{1/2\})\subset\mathbb{R}^2$. This set has open slices, but it is not open in the product topology (it contains $(0,0)$ but does not contain any ball around $(0,0)$ since $f(a,a)=1/2$ for any $a\neq 0$). Even more explicitly, this set $U$ is the complement of the set $\{(a,a):a\neq0\}$, which you can easily verify has closed slices but is not closed


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.