I just read the fact that open subspace of compact Hausdorff space is locally compact.
I wanted to know that is this true or false if we relax Haudroff condition?
I followed following defination of locally compact space at a.
A is locally compact space at a if there is compact subspace of A which contain non-empty nbhd of a.If space is locally compact at every point then A is locally compact space
Please Help me to construct counterexample or proof
Any Help will be appreciated
Let us call a space $Z$ locally compact if for each $z \in Z$ and each open neighborhood $U$ of $z$ in $Z$ there exists a compact neighborhood $C$ of $z$ in $Z$ such that $C \subset U$. Call $Z$ weakly locally compact if each $z \in Z$ has a compact neighborhood.
It seems that you ask for "weakly locally compact".
For non-Hausdorff spaces the answer is no. Take any not weakly locally compact space $X$ (for example $X = \mathbb Q$ with the subspace topology inherited from $\mathbb R$). Let $Y = X \cup \{*\}$ with a point $*$ not belonging to $X$. Define a topology on $Y$ as follows: $U \subset Y$ is open if either $U = Y$ or $U$ is an open subset of $X$. The verification that this is a topology on $Y$ is easy. Clearly $X$ is open in $Y$. Moreover, $Y$ is compact because each open cover of $Y$ must contain $Y$.
