$\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $. Does this implies $(Y, \tau_Y) $ is compact?

Question

$(X, \tau_X) $ and $(Y, \tau_Y) $ be two topological spaces.

$\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $.

Question : Does this implies $(Y, \tau_Y) $ is compact?

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Notation:

$Y^X$: Set of all functions from $X$ to $Y$.

$C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$

$\text{Gr}(f) =\{(x,f(x)):x\in X\}\subset X×Y$

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I want to show if $Y$ is not compact, then $\exists f\in Y^X$ with $\text{Gr}(f) $ is closed but $f\notin C(X, Y) $

If we choose $X=Y=\Bbb{R}$ and endowed with euclidean topology, then

$f(x)= \begin{cases}\frac{1}{x} & x\neq 0 \\ 0 & elsewhere \end{cases}$

is one such example of required functions.

But how to prove for any two topological spaces $X$ and $Y$ ?

A variant of this interesting question is on my MO post.

Answer 1

If $X$ has discrete topology then every function: $f:X \to Y$ is continuous. In particular, any function with closed graph is continuous. But $Y$ can be any topological space, not necessarily compact.

Answer 2

In general I think the following would work, let $Y$ be a non-compact topological space in which there is a distinguished point $y_0$ such that for every point $y$, if $V$ is an open neighbourhood of $y$, then $y_0 \in V$. Moreover suppose this distinguished point $y_0$ has a decreasing sequence of open neighbourhoods $U_n$ such that $\cap U_n = \{y_0 \}$.

Suppose $f : X \rightarrow Y$ has a closed graph, $X \neq \emptyset$, and assume $f(X) \neq \{y_0 \}$. Then there is some point $x \in X$ such that $f(x) = y_1 \neq y_0$. Consider the constant sequence $x_n = x$ in $X$, which converges to $x$ in $X$.

Then clearly $f(x_n) \rightarrow y_1 = f(x)$ because the sequence is constantly equal to $y_1 = f(x)$. But if $V$ is any open neighbourhood of $y_1$, then $y_0 \in V$, hence $f(x_n) \rightarrow y_0$ as well.

But by construction, $x_n \rightarrow x$ and $f$ has a closed graph, so we must have $f(x) = y_1$ and $f(x) = y_0$ which is absurd because $y_1 \neq y_0$, therefore if $f$ is not the constant function $X \rightarrow y_0$, $f$ does not have a closed graph (unless it is the empty function).

Now since the constant function $X \rightarrow y_0$ is continuous (as is the empty function), the space $Y$ satisfies the condition 'For all $f \in Y^{X}$, if $\mathcal{G}(f) \subset X \times Y$ is closed then $f \in C(X,Y)$' for all topological spaces $X$, and it does so in a non-vacuous way because the constant function $X \rightarrow y_0$ does have a closed graph, since because $\cap U_n = \{y_0 \}$, any net $f(x_{\alpha}) = y_0$ can only converge to $y_0$, since for any point $y \neq y_0$, there is some open neighbourhood of $y_0$ in which $y$ is not contained, so the only (non-empty) function for which the closed graph property holds is the constant function $X \rightarrow y_0$, and this function is continuous.

So it would suffice to construct a space that is a candidate for $Y$.

One example I think works as a candidate is the following space:

Let $Y = (0,1]$ as a set, and let $\tau_{(0,1]}$ be the topology generated by the left half-open intervals $(a,1]$ where $a \in (0,1]$, i.e. the open sets consist of the whole space, the empty set, and all left half open intervals $(a,1]$.

This space is non-compact, since $(\frac{1}{n},1] , n \geq 2$ is an open cover, but any finite subcover omits a euclidean neighbourhood of $0$.

Moreover for any point $a \neq 1$, any open neighbourhood of $a$ must be of the form $(b,1]$ with $0 < b < a$, so that $1$ is contained in every open neighbourhood of a point $a \neq 1$.

Finally setting $U_n = (1 - \frac{1}{n},1]$, which are each open neighbourhoods of $1$, we have that $\bigcap_{n \geq 1} U_n = \{1 \}$, so $((0,1],\tau_{(0,1]})$ is a candidate for $(Y,\tau_{Y})$.