2

Let $X$ and $Y$ be metric spaces and Y be a compact space. To show that a function $f : X \to Y$ is continuous iff its graph $G(f) = \{ (x,f(x)) \in X \times Y \}$ is closed in $X \times Y$.

I have done the part that if $f$ is continuous then its graph $G(f) = \{ (x,f(x)) \in X \times Y \}$ is closed in $X \times Y$.

But having difficulty in the other direction that if its graph $G(f) = \{ (x,f(x)) \in X \times Y \}$ is closed in $X \times Y$ then $f : X \to Y$ is continuous.

Is $Y$ required to be Hausdorff for graph $G(f) = \{ (x,f(x)) \in X \times Y \}$ is closed in $X \times Y$ assuming that $f$ is continuous?

User8976
  • 12,966
  • 10
  • 45
  • 118

2 Answers2

2

HINT: If $f$ is not continuous, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converging to some $x\in X$ such that $\langle f(x_n):n\in\Bbb N\rangle$ does not converge to $f(x)$. Use this to show that $G(f)$ is not closed; you’ll need to use the compactness of $Y$.

Error 404
  • 6,046
Brian M. Scott
  • 631,399
  • One doubt; Since $Y$ is compact so $f(x_n)$ has a convergent subsequence say $f(x_{n_k})$ converging to some $y$. Using the fact that $G(f)$ is closed it is evident that $f(x)=y$ How to derive a contradiction from here?? – Learnmore Dec 13 '17 at 13:55
2

First note why $Y$ being compact is necessary. The graph of $$f(x)=\begin{cases}0&x=0\\\frac{1}{x}&x\neq 0\end{cases}$$

is closed in $\mathbb R\times \mathbb R$.

Now, assume $Y$ is compact.

Let $x_n\to x$ and assume $f(x_n)$ does not converge to $f(x)$. But $f(x_n)$ is an infinite sequence in compact $Y$, so it must have a convergent subsequence to some $y\neq f(x)$. But then $(x_n,f(x_n))$ must have a convergent subsequence converging to $(x,y)$. Since $G$ is closed, then $(x,y)\in G$. But $(x,y)\in G$ iff $y=f(x)$.

Thomas Andrews
  • 186,215