A topology exercise about compactness and Hausdorff Let $0^{\prime}$ denote a point that is not an element of $(-1, 1)$, and let $X = (-1,1)\cup\lbrace 0^{\prime} \rbrace$. Let $X$ be equipped with the topology generated by $(-1,a), (a,1), (-1,b)\setminus \lbrace 0 \rbrace \cup \lbrace 0^{\prime} \rbrace, (c,1) \setminus \lbrace 0\rbrace \cup \lbrace 0^{\prime} \rbrace$, where $-1<a<1, 0<b<1, -1<c<0$. (We can think of $X$ as $(-1,1)$ with $0$ split in two. (Folland 4.37)
(a) Show that $X$ is not Hausdorff
(b) Show that $[-1/2,1/2]$, $[-1/2,1/2]\setminus \lbrace 0\rbrace \cup \lbrace 0^{\prime} \rbrace$ are compact but not closed in $X$ and their intersection is not compact.
I think for (a) $0$ and $0^{\prime}$ cannot but separated by open sets but I have no idea how to formally prove it. For (b) I'm trying to use the finite open subcover definition to show the compactness but that approach didn't quite work. The non-closedness also seems intuitively true but I'm not sure how to formally prove it.
Does anyone have solutions or ideas? Thanks.
 A: Your suspicion that $0$ and $0'$ can’t be separated by disjoint open sets is correct. Here’s a straightforward approach. Suppose that $U$ is an open nbhd of $0$ and $V$ an open nbhd of $0'$. Show that there are $a\in(-1,0)$ and $b\in(0,1)$ such that $(a,0)\cup\{0\}\cup(0,b)\subseteq U$ and $(a,0)\cup\{0'\}\cup(0,b)\subseteq V$. Conclude that $U\cap V\ne\varnothing$.
To show that such $a$ and $b$ exist, you could show that every intersection of finitely many members of the subbase given in the statement of the problem is equal to an intersection of just two members of it. Every such intersection that contains $0$ or $0'$ has one of the two forms that I gave above, so those sets are bases at $0$ and $0'$. In fact you’ll find that these two-fold intersections give you a base of open intervals, the only complication being that if $a<0<b$, then there are two $(a,b)$ intervals, one containing $0$ but not $0'$, the other containing $0'$ but not $0$. Thus, $X\setminus\{0'\}$ has exactly its usual topology, and $X\setminus\{0\}$ is identical apart from the name given to the point in the middle.
Now let $K=\left[-\frac12,0\right)\cup\{0\}\cup\left(0,\frac12\right]$ and $K'=\left[-\frac12,0\right)\cup\{0'\}\cup\left(0,\frac12\right]$; you’re to show that $K$ and $K'$ are compact but not closed in $X$. Once you’ve done the work for (a), showing that $K$ and $K'$ aren’t closed is easy: $0'$ is a limit point of $K$, and $0$ is a limit point of $K'$. The argument for compactness is the same for both sets. Given an open cover $\mathscr{U}$ of $K$, say, there must be some $U_0\in\mathscr{U}$ that contains $0$. The topology on $K\setminus U$ is identical to the topology on $\left[-\frac12,\frac12\right]\setminus U$ in $\Bbb R$, so $K\setminus U$ is compact, and only finitely many members of $\mathscr{U}$ are required to cover it.
That’s really the whole argument; what I’ve left for you are the details needed for the results in the second paragraph; that’s really the heart of both parts.
