Some topological property of a quotient space Consider a plane with y-axis identified equipped with quotient topology,then whether this space is Hausdorff and locally compact or not?
Intuitively,we can find a strip as the neighborhood of the y-axis which has been identified.But the "strip" can separate the "y-axis" point with other point on the plane,my guess is it's Hausdorff. Additionally since this strip is always infinte, we couldn't find a compact neighborhood of "y-axis" point. My problem is that I don't know how to transfer these ideas into mathematical language ?
 A: Let $X$ be the quotient space, let $Y$ be the $y$-axis, and let $p\in X$ be the point corresponding to $Y$; we can identify $X\setminus\{p\}$ with $\Bbb R^2\setminus Y$, so it’s easy to check that distinct points of $X\setminus\{p\}$ have disjoint open nbhds. 
If $\langle x,y\rangle\in X\setminus\{p\}=\Bbb R^2\setminus Y$, let $a=\frac{x}2$; then $(\leftarrow,a)\times\Bbb R$ and $(a,\to)\times\Bbb R$ are disjoint open sets in $\Bbb R^2$, one containing $\langle x,y\rangle$ and the other containing $Y$. (Which one contains $Y$ depends on whether $,$ is positive or negative.) It’s easy to check that their images under the quotient map $q:\Bbb R^2\to X$ are disjoint open nbhds of $\langle x,y\rangle$ and $p$ in $X$, because they are saturated open sets: if $U$ is either of them, $q^{-1}\big[q[U]\big]=U$. (In case you’ve not encountered them before, $(\leftarrow,a)$ and $(a,\to)$ are another standard notation for $(-\infty,a)$ and $(a,\infty)$, respectively.) It follows that $X$ is indeed Hausdorff.
You are also correct in thinking that $X$ is not locally compact at $p$. Let $U$ be any open nbhd of $p$ in $X$. Let $V=q^{-1}[U]=Y\cup q^{-1}[U\setminus\{p\}]$; $V$ is an open nbhd of $Y$ in $\Bbb R^2$, so for each $n\in\Bbb N$ there is a point $\langle x_n,n\rangle\in V\setminus Y$. Let $D=\{\langle x_n,n\rangle:n\in\Bbb N\}$; $D$ is an infinite, closed, discrete subset of $\Bbb R^2$. To complete the argument, show that $D$ is also a closed, discrete subset of $X$. (The hardest part of this is showing that $p\notin\operatorname{cl}_XD$, which amounts to showing that $Y$ has an open nbhd in $\Bbb R^2$ that is disjoint from $D$.) It then follows that $D$ is a non-compact closed subset of $\operatorname{cl}_XU$ and hence that $\operatorname{cl}_XU$ is not compact.
