Compact subset in two different topologies

Question

Let $(X,\tau_1)$ and $(Y,\tau_2)$ be two Hausdorff topological spaces such that $X\subset Y$. Let $K\subset X$ be compact. Is $K$ also compact in $Y$? If not, what if both $X$ and $Y$ are compact spaces?

Answer 1

Compactness depends on the topology, so if $X\subset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=\mathbb{R}$ with Euclidean topology and $Y=\mathbb{R}$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).

Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $X\cap U$ with $U\in\tau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.

Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).

Answer 2

If $K\subseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.

So if also $K\subseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $X\subseteq Y$).

Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.