Suppose that $(X,T)$ is a non-compact Hausdroff space and $T'=\{\varnothing\}\cup\{X-K: K$ is $T$-compact$\}$.
Would $X$ equipped with $T'$ have to be a second countable space? Or at least first countable? How about separability? Also, is it locally compact?
My thoughts on this:
I found that $(X,T')$ is compact and that every closed subset of it is also compact. It's non-Hausdorff and weaker than $T$.
I did the most natural thing taking $X = \Bbb R$ as a start. I found this to be locally compact, I'm not sure though of its other properties. It's claimed somewhere on the Internet that it is in fact second countable.
So the example of $\Bbb R$ is not much insightful. I am having hard time thinking of a suitable example that would disprove local compactness, let alone the other properties.
Thanks for your help.
Let $\langle X,\tau\rangle$ be an uncountable discrete space. Then the only $\tau$-compact subsets of $X$ are the finite sets, and $\langle X,\tau\rangle$ is certainly locally compact, non-compact, and Hausdorff. But $\tau'$ is just the cofinite topology, which certainly isn’t second countable.
Added: The construction of $\tau'$ from a locally compact, non-compact, Hausdorff topology $\tau$ on some set $X$ will always result in a compact space. Suppose that $\mathscr{U}$ is a $\tau'$-open cover of $X$. Let $\mathscr{F}=\{X\setminus U:U\in\mathscr{U}\}$; then $\mathscr{F}$ is a family of $\tau$-compact sets with empty intersection, so there must be a finite $\mathscr{U}_0\subseteq\mathscr{U}$ such that $\bigcap_{U\in\mathscr{U}_0}(X\setminus U)=\varnothing$ and therefore $\bigcup\mathscr{U}_0=X$.
Added2: Let $\tau$ be the usual order topology on $X=\omega_1$; $\langle X,\tau\rangle$ is locally compact, non-compact, and Hausdorff, and its compact sets are precisely those that are closed and bounded in the usual order. Thus, the non-empty members of $\tau'$ are the $U\in\tau$ for which there is an $\eta<\omega_1$ such that $\{\xi\in X:\eta<\xi\}\subseteq U$. If $D\subseteq X$ is countable, then $\{\xi\in X>\sup D<\xi\}$ is a $\tau'$-open set disjoint from $D$, so $\langle X,\tau'\rangle$ is not separable.
Added3: For completeness I’m adding the result (from the comments below) that $\tau'$ is always a locally compact topology if $\tau$ is. We first show that if $C\subseteq X$ is $\tau$-closed, then $C$ is $\tau'$-compact. Let $\mathscr{U}$ be a $\tau'$-open cover of $C$, and for each $U\in\mathscr{U}$ let $K_U=X\setminus U$; $K_U$ is $\tau$-compact. Fix $V\in\mathscr{U}$; then $C\cap K_V$ is $\tau$-closed and hence $\tau$-compact. Since $\tau'\subseteq\tau$, $\mathscr{U}\setminus\{V\}$ is a $\tau$-open cover of $C\cap K_V$, so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}\setminus\{V\}$ such that $C\cap K_V\subseteq\bigcup\mathscr{U}_0$. But then $\mathscr{U}_0\cup\{V\}$ is a finite subset of $\mathscr{U}$ that covers $C$, which is therefore $\tau'$-compact.
Now let $x\in X$, and let $U$ be a $\tau'$-open nbhd of $x$. Let $K=X\setminus U$. Each $y\in K$ has a $\tau$-open nbhd whose $\tau$-closure is $\tau$-compact and does not contain $x$, and $K$ is $\tau$-compact, so $K$ has a $\tau$-open nbhd $V$ whose $\tau$-closure is $\tau$-compact and does not contain $x$. Let $W=X\setminus\operatorname{cl}_\tau V$; then $W$ is a $\tau'$-open nbhd of $x$, and $\operatorname{cl}_\tau W\subseteq X\setminus V\subseteq X\setminus K=U$, so $\operatorname{cl}_\tau W$ is a $\tau'$-compact $\tau'$-nbhd of $x$ contained in $U$, and $\langle X,\tau'\rangle$ is locally compact.