Question : Let $\Omega_1 \subset \Omega_2$ be two topological structures in $X$. Does the compactness of $(X, \Omega_2)$ imply that of $(X, \Omega_1)$? And vice versa?
My attempt : Assuming the space is compact wrt. $\Omega_1$ we can find a finite subcover for open sets in $\Omega_1$ then these sets are also in $\Omega_2$ thus the space is compact in $\Omega_2$ BUt this argument is wrong. What am i missing ?
On
Suppose $X$ is compact wrt $\Omega_2$. We need to show that it is compact wrt $\Omega_1$. Let $\{U_\alpha\}$ be an open cover of $X$ in $\Omega_1$. That is each $U_\alpha$ is in $\Omega_1$. This means each $U_\alpha$ is in $\Omega_2$ and $X$ is compact in $\Omega_2$ so we have a finite subcover $U_{\alpha_1},\cdots ,U_{\alpha_n}$ of $X$ which is what we needed.
Conversely, if $X$ is compact in $\Omega_1$ it need not be compact in $\Omega_2$. Consider for example $\Omega_1$ is the indiscrete topology and $\Omega_2$ is the discrete topology on $X$ and $X$ is an infinite set.
Assume $X$ compact with respect to $\Omega_1$. Choose an arbitrary open cover with respect to $\Omega_2$. Is this an open cover with respect to $\Omega_1$ too, in order for you to be able to extract an open subcover with respect to $\Omega_1$? No, because open sets from $\Omega_2$ do not necessarilly belong to $\Omega_1$.
The converse, though, is true. Assume $X$ compact with respect to $\Omega_2$. Choose an arbitrary open cover $\Omega_1$. This is still an open cover with respect to $\Omega_2$. Pick here an finite subcover. This will also be an open subcover with respect to $\Omega_1$, because the open subsets were originally from $\Omega_1$. Therefore $X$ is compact with respect to $\Omega_1$.