Graph of f closed and f is continuous

Question

Given a function f:X→Y, show that if X is Hausdorff and Y is compact, then f is a continous function if and only if G(f) or graph of f, is a closed subset of X×Y. I could only prove graph is closed in case Y is Hausdorff and f is continuous. It would be helpful if someone could tell how to do this. Can we do it using the notion of nets? Sorry, I'm new to topology.

Answer 1

Suppose $G(f) \subseteq X \times Y$ is closed. Suppose that $Y$ is compact.

A useful fact (Kuratowski's theorem) is that $\pi_X: X \times Y \to X$ then is a closed map, for any $X$. See this question, e.g. and its answers.

$f$ is continuous iff for every closed set $C \subseteq Y$, $f^{-1}[C]$ is closed in $X$.

Now, for such a closed $C$ we can say:

$$f^{-1}[C]=\pi_X[(X \times C) \cap G(f)]$$

(Proof: $x \in f^{-1}[C]$ then $f(x) \in C$ so $(x,f(x)) \in G(f) \cap (X \times C)]$ and $x=\pi_X(x,(f(x))$ so $x$ is in the right hand set. Conversely, $x \in \pi_X[(X \times C) \cap G(f)]$ if $\exists y: (x,y) \in (X \times C) \cap G(f)$ but that forces $y=f(x)$ and $y \in C$ so $f(x) \in C$ or $x \in f^{-1}[C]$)

Now $X \times C$ is closed in $X \times Y$ so the intersection with $G(f)$ is also closed as the intersection of two closed sets, and then the fact that $\pi_X$ is a closed map tells us that $f^{-1}[C]$ is thus closed, as required. So $f$ is continuous.

Answer 2

To make some other points and additions:

The other direction is false as stated. From $X$ Hausdorff and $Y$ compact and $f$ continuous we cannot conclude $G(f)$ is closed in $X \times Y$:

Let $X=\Bbb R$ usual topology, $Y=\Bbb R$, indiscrete (trivial) topology, $f: X \to Y: f(x)=x$ is continuous (any function to an indiscrete space), $Y$ is trivially compact, and $X$ is Hausdorff (being even metric). But $G(f)$ is dense in $X \times Y$: if $(x,y) \in X \times Y$, a basic neighbourhood of it looks like $O \times Y$, where $O$ is a usual open neighbourhood of $x$ in $X$, and then $(x,x) \in G(f) \cap (O \times Y)$ so that $(x,y) \in \overline{G(f)}$.

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We can conclude that $G(f)$ is closed in $X \times Y$ from $Y$ being Hausdorff (as the only assumption):

If $f:X \to Y$ is continuous: $f \times 1_Y: X \times Y \to Y^2$ defined by $(f \times 1_Y)(x,y)=(f(x),y)$ is continuous (for any spaces) as $\pi_1 \circ (f \times 1_Y) =f$ and $\pi_2 \circ (f \times 1_Y) = 1_Y$ are both continuous and $\pi_1,\pi_2$ are the projections for $Y^2$ (universal mapping property for products).

$Y$ is Hausdorff iff $\Delta_Y:=\{(y,y): y \in Y\} \subseteq Y^2$ is closed in $Y^2$.

Finally $G(f)=(f \times 1_Y)^{-1}[\Delta_Y]$

($(x,y)$ is in this set iff $(f \times 1_Y)(x,y) \in \Delta_Y$ iff $(f(x),y) \in \Delta_Y$ iff $y=f(x)$ iff $(x,y) \in G(f)$)

and so $G(f)$ is closed as the inverse image of a closed set under the continuous $f \times 1_Y$.

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An alternative proof for $X$ any space, $Y$ compact, $G(f)$ closed in $X \times Y$ then $f$ continuous, using nets:

Suppose $f$ is not continuous, then there is some $x \in X$ and a net $(x_a)_{a \in A}$ that converges to $x$ but such that $f(x)$ is not a cluster point of $(f(x_a))_{a \in A}$.

(Proof sketch: there is some open $O \subseteq Y$ such that some $x \in f^{-1}[O]$ is not an interior point of $f^{-1}[O]$, so using $A=\mathcal{U}(x)$ under reverse inclusion and picking $x(U) \in U \setminus f^{-1}[O]$ for each $U \in A$, we can easily see we have a net as required.)

Then $(x_a, f(x_a))_{a \in A}$ is a net in $G(f)$ and $(f(x_a))_{a \in A}$ is a net in the compact $Y$, so we have some $y_0 \in Y$ and a convergent subnet $(f(x_b))_{b \in B}$ that converges to $y_0$ by compactness. As $(x_b, f(x_b))_{b \in B}$ converges to $(x,y_0)$ and as $G(f)$ is closed, $y_0=f(x)$ but this contradicts the choice of the net in that $f(x)$ then is a cluster point of $(f(x_a))_{a \in A}$. So $f$ is continuous.

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Hopefully these remarks help some people.