I was asking this question at [Mathematics SE](https://math.stackexchange.com/questions/4128343/translating-limit-point-under-homeomorphism-map) 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.



Limit points and Homeomorphism
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 $(h\circ g)(x\_n)\to 0$.


