$T_1$ property and local compactness Recall that a space  $X$ is a locally compact if for every $x\in X$ there exists a neighbourhood $U$ of the point of $x$ such that $cl U$  is compact subspace of $X$. How we can show that  compact $cl U$ is $T_1$?
 A: $\newcommand{\cl}{\operatorname{cl}}$Engelking defines compact to mean what I would call compact and Hausdorff. It is true that if each point of a space $X$ has a nbhd $U$ such that $\cl U$ is compact and Hausdorff, then $X$ is $T_1$. More generally, if $X$ is any space, and $x\in X$ has a nbhd $U$ such that $\cl U$ is compact and Hausdorff, then $\{x\}$ is closed in $X$. (Recall that a space is $T_1$ iff each singleton set is closed.)
To see this, let $x\in X$, and let $U$ be a nbhd of $x$ such that $\cl U$ is compact and Hausdorff. The space $\cl U$ with the relative topology inherited from $X$ is a Hausdorff space, so it’s $T_1$, and therefore $\{x\}$ is a closed set in $\cl U$. This means that there is some closed set $F$ in $X$ such that $\{x\}=F\cap\cl U$. But then $\{x\}$ is the intersection of two sets that are closed in $X$, and as such it must itself be closed in $X$. 
A: $\mathbb{Z}$ with the digital line topology is a counterexample to this claim. Denote $\mathbb{Z}$ with the digital line topology as: $\mathbb{Z}_{\text{DL}}$ Note that every integer $m$ has a neighborhood contained in a compact subset of $\mathbb{Z}_{\text{DL}}$. If $m$ is odd take the neighborhood to be $\{m\}$ which is contained in $\{m,m+1,m+2\}$, a compact subset. If $m$ is even then take the neighborhood to be $\{m-1,m,m+1\}$ which is contained in $\{m-1,m,m+1\}$, a compact subset. 
Think about $2\in \mathbb{Z}$. A compact set containing a neighborhood of $2$ is: $\{1,2,3\}$ Yet $\{1,2,3\}$ as a subspace of $\mathbb{Z}_{\text{DL}}$ has the following open sets: $\{\{1\},\{1,2,3\},\{3\},\emptyset \} $. We see that $2$ is inextricably linked to $3$ and $1$. Therefore, for the two distinct points $2$ and $3$, there does not exist an open subset $U$ of $\{1,2,3\}$ such that $2\in U$ and $3 \notin U$ so $\{1,2,3\}$ can't be $T_1$. 
