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I was asking this question at Mathematics SE but I got nothing at all. This is why I am trying this site.

We consider the topology of the extended real line. Let $h\colon [-\infty,\infty]\to\Bbb R$ and suppose $(-\infty,0)$ and $(\infty,0)$ are limit point of the graph of $h$, that is, they are limit points of $\{(x,h(x))\colon x\in\Bbb R\}$. Now, let $g\colon (a,b)\to \Bbb R$ be a homeomorphism. Notice that $\lim\limits_{x \to a^{+}}g(x)=\infty$ or $-\infty$ and $\lim\limits_{x \to b^{-}}g(x)=\infty$ or $-\infty$. This is true for any homeomorphism. Now, consider $h\circ g\colon (a,b)\to \Bbb R$.

Claim: $(a,0)$ and $(b,0)$ are limit points of the graph of $h\circ g$. I think my claim would be true since homemomorphism behaves nicely with topological property.

My attempt was to consider a sequence $x_n$ in $(a,b)$ and show $(x_n, (h\circ g)(x_n))\to (a,0)$ but I could not finish. Maybe I should start direct with definition of limit points. This is why I am asking. Any help will be appreciated greatly.

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    $\begingroup$ Link to the post at m.se? $\endgroup$ Commented May 6, 2021 at 23:18
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    $\begingroup$ @00GB, yes, you should. Presumably it's math.stackexchange.com/questions/4128343/… ? $\endgroup$ Commented May 6, 2021 at 23:25
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    $\begingroup$ So, it seems you waited one day to post here. It's recommended to wait a week. Also, please edit a link to this question into the m.se question. $\endgroup$ Commented May 6, 2021 at 23:28
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    $\begingroup$ Isn't this just a matter of inserting things into the definitions? Let's say that $g$ is increasing. There are $y_n\to -\infty$ such that $h(y_n)\to 0$. Write $y_n=g(x_n)$. Then $x_n\to a$ since $g$ is a homeomorphism, and $(h\circ g)(x_n)\to 0$. $\endgroup$ Commented May 7, 2021 at 16:55
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    $\begingroup$ Yes, a homeomorphism on an interval is increasing or decreasing; wlog I assumed it to be increasing. $\endgroup$ Commented May 7, 2021 at 21:49

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A homeomorphism on an interval is increasing or decreasing; wlog I assume it to be increasing. There are $y_n \to -\infty$ such that $h(y_n)\to 0$. Write $y_n=g(x_n)$. Then $x_n = g^{-1}(g(x_n)) \to a$ since $g$ is a homeomorphism and $y_n\to -\infty$, and $(h\circ g)(x_n)\to 0$.

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