Is $( \mathbb{R}, T_{k}) $ locally compact, where $T_{k}$ is $K$-topology on $\mathbb{R} $? Is $( \mathbb{R}, T_{k}) $ locally compact, where $T_{k}$ is $K$-topology on $\mathbb{R} $?
I am stuck. Any hint helps.
P.s. $X$ is called locally compact, if every point $x \in X$ has a compact neighbourhood.
 A: No.
First of all note that for any $x\in K$ we have that any open neighbourhood of $x$ has to contain some interval $(x-\epsilon, x+\epsilon)$. This shows that $x$ belongs to the closure of $(x-\upsilon, x+\upsilon)-\{x\}$ for any $\upsilon>0$. And this is not a surprise, after all $K$-topology behaves like the euclidean topology around any point other than $0$.
Now let $U$ be an open neighbourhood of $0$. Then there is $\epsilon>0$ such that for $V:=(-\epsilon,\epsilon)-K$ we have $V\subseteq U$. From my first observation it follows that $\overline{V}$ contains all but finitely many elements of $K$. Therefore all but finitely many elements of $K$ are contained in $\overline{U}$. But $K$ has no limit points, in particular an infinite sequence composed of different elements of $K$ does not have a limit. Therefore $\overline{U}$ is not compact.
Side note 1: $0$ is the only point that behaves pathologically. Any other point has compact neighbourhood.
Side note 2: there are many nonequivalent definitions of local compactness but in Hausdorff space they are all equivalent. And $K$-topology is Hausdorff.
