Compactness of a set with respect to different spaces.

Question

Let $X$ be a topological space. $E\subseteq X$. Suppose $E$ is compact with respect to $X$, namely for every open covering $\{U_i\}_{i\in I}$ with respect to $X$ (I mean such open covering can exceed $E$, that is $E\subsetneq U_i$) of $E$, there exists $\{U_i\}_{i\in J:J~\text{:finite}}$ that covers $E$. Then if $E$ is thought of as a subspace of another arbitrary topological space $Y$, $E$ is compact with respect to such $Y$. I guess this is true, but is there a quick way to think or to show it?

Answer 1

Yes, it is true, because being compact only dependes upon the open subsets of $E$. In fact, if $E\subset X$ and if $\{U_\lambda\mid\lambda\in\Lambda\}$ is an open cover of $E$, in which each $U_\lambda$ is an open subset of $X$, then what really matters are the stes $U_\lambda\cap E$, which will form an open cover of $E$ such that each of its elements is an open subset of $E$. And assrting that this new cover will have a finite subcover is the same thing as asserting that the orifinal cover $\{U_\lambda\mid\lambda\in\Lambda\}$ has a finite subcover. In fact, if $F\subset\Lambda$,$$E\subset\bigcup_{\lambda\in F}U_\lambda\iff E=\bigcup_{\lambda\in F}U_\lambda\cap E.$$

Answer 2

It's not true, because $[0,1]$ is compact in $R$ standard topology but not in the lower limit topology, because $[0,1]$ is not countable (see https://en.m.wikipedia.org/wiki/Lower_limit_topology).

But if $Y$ is a subspace of $X$ or $X$ is a subspace of $Y$ it is always true.