Compact in one topology does not necessarily imply in other topology. Let $S$ be the set $$\mathbb{R}^2\setminus \{(0, y)\ |\ y\neq 0\}$$ obtained by removng the $y$-axis except for the origin. Let $\tau_1$ denote the subspace topology on $S$ induced from the usual topology of $\mathbb{R}^2$. Next, consider the surjective map $\pi:\mathbb{R}^2\to S$ defined by $$(x, y)\mapsto (x, y)\ \text{for}\ x\neq 0, \ \text{and}\ (0, y)\mapsto (0, 0)\ \text{for all}\ y.$$
Let $\tau_2$ be the resulting quotient topology on $S$. Then show that there is a subset of $S$ that is compact with respect to $\tau_2$ but not compact with respect to $\tau_1$.
I know that a set is compact if every open cover of it has a finite subcover. But I am not able to identify any such subset. Please help.
 A: Let
$$C=\{x\in S\vert d(0,x)\le 1\}$$
this set is closed and bounded and, hence, compact. Whereas
$$\pi^{-1}(C)= \{x\in S\vert d(0,x)\le 1\}\cup \{(0,y)\vert y\in\mathbb{R}\}$$
is not bounded and, hence cannot be compact in $\mathbb{R}^2$
A: Let $A=\{\,(x,y)\,|\,\, x^2+y^2\leq1 \text{$\,\,\,$and$\,\,\,$} (x,y)\in S\,\}$. Then $A$ is compact with respect to $\tau_2$ but not with respect to $\tau_1$.
My reasoning:

*

*$A$ is not open in usual topology, so it is not compact. Since, $\tau_1$ is usual topology. Furthermore, we can also imagine an open covering which has no sub-covering.


*Key idea: "Any open set in $\tau_2$ which contains the origin $(0,0)$, must contain a strip like $$U_{\epsilon}=((-\epsilon,\epsilon)\times\Bbb{R}^1)\cap S$$ due to the definition of the quotiend topology." So, given an open covering of $A$ in $\tau_2$, it contains an open set $U$ which contains such an $U_{\epsilon}$. Then consider the closed and bounded set $B=A-U_{\epsilon}$. It does not contain $(0,0)$, so for it the topology is the usual one. So we can find a finite subcover for it from the given covering. This subcover together with $U$ is a subcovering for $A$.
I need to improve my reasoning. I hope my example is correct.
