# For every $U \subseteq V$, $U$ is compact in $V$ if it is compact in $X$, or, for every $U \subseteq V$, $U$ is compact in $X$ if it compact in $V$.

Given:

$$(X, \tau)$$ is a topological space and $$V \subseteq X$$ is a closed subset that is equipped with the subspace topology.

For each of the following statements. Prove that it is either true or false. If it is false, provide a counter example.

1. For every $$U \subseteq V$$, $$U$$ is compact in $$V$$ if it is compact in $$X$$.
2. for every $$U \subseteq V$$, $$U$$ is compact in $$X$$ if it compact in $$V$$.

Attempt at solution:

If $$U$$ is compact in $$V$$ then there is an open cover $$\{C_{i}\}$$ of $$U$$ with respect to $$X$$. Then $$\{C_i\cap V\}$$ is an open cover of $$C$$ with respect to $$V$$. So it has a finite subcover $$\{C_1\cap V,\ldots, V_n\cap V\}$$. Meaning that $$\{C_1,\ldots, C_n\}$$ is an open subcover of $$\{C_i\}$$. Thus $$U$$ is compact with respect to $$X$$.

• attempt added. @JoséCarlosSantos – J. Watson Oct 17 '18 at 6:53

You should start by saying 'let $$\{C_i\}$$ be an open cover' instead of 'there is an open cover'. Except for this your proof is correct. For the other part let $$\{C_i\}$$ be an open cover of $$U$$ in the topology of $$V$$. Then we can write $$C_i=V_i\cap V$$ for some open sets $$V_i$$ in $$X$$. Hence $$U \subset \cup (V_i\cap V) \subset \cup V_i$$. Since $$U$$ is compact in $$X$$ there is a finite subcover, say $$V_1,V_2,...,V_k$$. Now use the fact that $$U \subset V$$ to conclude that $$C_1=V_1\cap V,C_2=V_2\cap V,...,C_k=V_K\cap V$$ is a subcover in $$V$$.