If each slice of $f : X \times Y\rightarrow Z$ is continuous then $f$ is continuous?

Question

I'm having trouble to conclude my proof of the next statement.

If $f:X \times Y \rightarrow Z$ is a function such that, $f_x : Y \rightarrow Z$ given by $f_x(y) = f(x,y)$ is continuous for each $x \in X$ and $f_y : X \rightarrow Z \ $ given by $\ f_y(x) = f(x,y)$ is continuous for each $y \in Y\ $ then $f$ is continuous

This is what I've tried, given $(x,y) \in X \times Y$ and $\ U \in \tau_Z$ such that $f(x,y) \in U$ we take $\ f_x, \ f_y$ continuous functions, so $f_x^{-1}(U) \in \tau_Y \text{ and } f_y^{-1}(U) \in \tau_X$ so we take $V = f_y^{-1}(U) \times f_x^{-1}(U)$ wich is open in $X \times Y$ and $(x,y) \in V$.

My problem is when I have to show that $f(V) \subseteq U$. I need some help

Edit: I've saw something similar here and uses strongly compactness, so this is likely false, But I don't know how can be a function wich each are continuous but not the whole function.

Answer 1

This is not correct in general. Take for example the function $f = \begin{cases} \frac{xy^2}{x^2+y^4} \ , \quad &(x,y)\neq (0,0) \\ 0 \ , \quad &\text{otherwise} \end{cases}$.

Then, $f_x$ is continuous for every $x \in \mathbb{R}$ and $f_y$ is continuous for every $y \in \mathbb{R}$. But $f$ fails to be continuous at $(0,0)$, since $f(x,0)=0 \to 0$ but $f(y^2,y)=\frac{1}{2} \to \frac{1}{2}$ (as $(x,y)\to (0,0)$).

Answer 2

At least in the context of $X=Y=[0,1]$ and $Z=\mathbb{R}$, this is essentially the same as

If $(g_n)_{n\geq 1}$ is a sequence of continuous functions $[0,1] \to \mathbb{R}$ that converges pointwise to the continuous function $g$, then $g_n$ converges to $g$ uniformly.

which is false. For instance, we may consider the sequence $$g_n(x) = \begin{cases}nx(1-nx) & 0 \leq x < \frac{1}{n} \\ 0 & \text{ otherwise}\end{cases}$$ which converges to $g(x) = 0$ pointwise, but $g_n(1/(2n)) = \frac{1}{4} \not\to 0,$ so $g_n$ doesn't converge to $g$ uniformly.

For this particular example, we may create our $f$ by formally replacing $n$ with $1/y$ to get $$f(x,y) = \begin{cases}\frac{x}{y}\left(1-\frac{x}{y}\right) & 0 \leq x < y \\ 0 & \text{ otherwise}\end{cases}.$$

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In general, if we have a sequence $g_n \to g$ pointwise but not uniformly, then we define $f : [0,1]^2 \to \mathbb{R}$ by $$f(x,y) = \begin{cases}\left(n+1-\frac{1}{y}\right)g_n(x) + \left(\frac{1}{y} - n\right)g_{n+1}(x) & \text{ if } y > 0, n=\lfloor 1/y\rfloor \\ g(x) & \text{ if } y=0\end{cases}$$

Since the functions $g$ and $(g_n)_{n\geq 1}$ are continuous, each $f(\cdot, y)$ is continuous. Since $g_n \to g$ pointwise, each $f(x,\cdot)$ is continuous.

However, one may use the compactness of $X=[0,1]$ along with the assumption $g_n$ doesn't converge to $g$ uniformly to show that $f$ is not continuous at some point $(x,0)$.

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Conversely, if $f : [0,1]^2 \to \mathbb{R}$ were a function with continuous slices but not continuous at $(a,b)$, then there is a sequence of points $(a_n,b_n) \in [0,1]^2$ such that $(a_n,b_n) \to (a,b)$ but $f(a_n,b_n) \not\to f(a,b)$. Then we may define $$g_n(x) = f(x,b_n) \\ g(x) = f(x,b).$$ Once again, since $f$ had continous slices, these functions are continuous and $g_n \to g$ pointwise, but $g_n(a_n) \not\to g(a)$, so the convergence is not uniform.