Let $f:X\to Y$ be a function in the metric spaces $X,Y$. The graph of $f$ is the set: $$G(f)=\{(x,f(x));x\in X\}$$ And we say that the graph of $f$ is closed, if the set $G(f)$ is closed in $X\times Y$ by product topology.
Now consider the function $f:[-\frac{2}{\pi},\frac{2}{\pi}] \to [-1,1]$ by: $$ f(x) =
\begin{cases}
\sin(\frac1x), & \text{if $x\ne 0$ } \\
0, & \text{if $x=0$}
\end{cases}$$ My question is this: Is the graph of $f$ closed? I think the answer is no. Since if the graph of $f$ is closed, it is bounded in $\mathbb R^2$ and so, should be compact. But if $G(f)$ is compact and the domain of $f$ be compact, then $f$ is continuous (see here). Clearly $f$ isn't continuous at $x=0$. It's enough to consider the sequence $x_n=\frac{1}{n\pi}$ and $w_n=\frac{1}{\frac{\pi}{2}+2n\pi}$. Then, $\lim x_n=\lim w_n =0$ as $n\to \infty$ but $\lim f(w_n)=1$ and $\lim f(x_n)=0$ as $n\to \infty$.
Is my analysis correct?
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hamid kamali
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2You may see this directly: you have a sequence of points $(w_n,f(w_n))$ that converges to the point $(0,1)$ which is not in $G$. – Jean-Claude Arbaut Nov 06 '15 at 17:19
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1Regarding your idea with continuous functions, you are right, see this blog post by Terence Tao (look for Theorem 5 (Closed graph theorem (point set topology))) – Jean-Claude Arbaut Nov 06 '15 at 17:29
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@Jean-ClaudeArbaut thank you for introducing the book and your point in first comment. – hamid kamali Nov 06 '15 at 17:46
1 Answers
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Hint.
You're right the graph $G(f)$ is not closed. The point $(0,1)$ belongs to the complement of $G(f)$ in $\mathbb R^2$. However, you can prove that a sequence of point of the graph converges to $(0,1)$.
mathcounterexamples.net
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