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I'm looking for whether the graph of topologist's sine curve and closed topologist's sine curve are closed or not. But due to some misconception, I'm facing problems with this.

$\underline{\text{Question} : 1}$ Here it proves that if $f\colon X\to Y$ continuous and $Y$ is Hausdorff, then the graph $G_f$ of $f$ is closed.

$f(x)=\sin\frac1x,x\in(0,1]$ is continuous and $\mathbb R^2$ is Hausdorff, this means $G_f$ is closed, but $G_f\ne\bar{G_f}$, implying $G_f$ is not closed.

$\underline{\text{Question} : 2}$ Here it proves that the graph $G_f$ of $f\colon X\to Y$ is closed in $X\times Y$, then $f$ is continuous if $Y$ is compact.

The closed topologist's sine curve is closed in $\mathbb R^2$. If we take a compact subset of $\mathbb R^2$ containing this graph, for example take $[-\frac12,1]\times[-2,2]$, then the whole curve lies there. And hence $f\colon[0,1]\to Y$ is continuous, which is surely not.

As a beginner in topology, I'm certainly missing something, but can't able to see what. It would be great if someone please point out where I'm going wrong.

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Question 1: what do you take $X$ here to be? If $X$ is $\Bbb{R}$, then $f$ is not defined on all of $X$ (it is only defined in $(0,1]$). IF $X$ is $(0,1]$, then the graph should be considered as a subset of (say) $(0,1]\times \Bbb{R}$. And there it is indeed a closed subset.

Question 2: The "The closed topologist's sine curve" is not a graph of any function: there is more than one $y$ with $(0,y)$ in it. So this result cannot be applied.

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For $f(x)=\sin(\frac{1}{x})$ with domain $(0,1]$, we do have that $G_f$ is closed as a subset of $(0,1]\times [0,1]$ which is consistent with the continuity of $f$ on that domain.

If we extend $f$ to a function on $[0,1]$ then whatever value $t\in [-1,1]$ we use for $f(0)$, the resulting $f$ has a non-closed graph $G_f$ in $[0,1] \times [-1,1]$, as $\{0\} \times [-1,1]$ is in the closure of $G_f$, whatever $f(0)$ might be, again consistent with the non-continuity of an extended $f$ on $[0,1]$.

The total "topologist's sine curve", which includes $\{0\} \times [-1,1]$, is compact, connected, but is not the graph of any function, so we cannot apply a closed graph theorem to it, directly.

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