Suppose that we have two topological spaces $(X,T_1)$, $(X,T_2)$ such that $T_1\subset T_2$ and $K\subset X$ is compact in $T_1$ topology. Is it also compact in $T_2$ topology ???
I think that the answer is yes, because we can cover $K$ with a finite collection of open sets that belong in $T_1$ topology, but we know that these open sets belong also in $T_2$, so it is going to be compact in $T_2$.
Is this true ??
There is a theorem which you might have met already which answers your question fairly definitively:
If a topology is both compact and Hausdorff, then it is maximal as a compact topology and minimal as a Hausdorff topology. That is any strictly larger topology is not compact and any strictly smaller topology is not Hausdorff.
Moreover, this follows from this theorem:
Suppose $f: X \to Y$ is is a continuous bijection, $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.
Moreover, this follows from two even more basic results:
A compact subset of a Hausdorff space is closed, and a closed subset of a compact space is compact.
and
The continuous image of compact set is compact.
No, but it does go the other way.
Consider an extreme example. Consider the set $\mathbb{N}$ with two different topologies. First, let $T_1 = \{ \emptyset , \mathbb{N} \}$, the "trivial topology", and let $T_2 = \mathcal{P}( \mathbb{N} )$ be the discrete topology on $\mathbb{N}$. Then $\mathbb{N}$ is compact in $T_1$, but not in $T_2$. Specifically, consider the open cover $\mathcal{U} = \{ U_k \}_{k \in \mathbb{N}}$, where $U_k = \{ k \}$. Then $\mathcal{U}$ is an open cover of $\mathbb{N}$ with no finite subcover.
We can consider $X=[0,1]$, with $T_1$ standard, and $T_2$ discrete and $K=X$. Certainly, $T_1 \subset T_2$, but $[0,1]$ is not compact in the $T_2$ topology.
On the other hand, suppose that $T_1 \subset T_2$, and $K$ is compact wrt $T_2$. Let $U_\alpha$ be some open cover of $K$ in $T_1$, then, since $U_\alpha$ are open in $T_2$ as well, you can find a finite subcover, so it is compact in $T_1$ as well.
