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 $V$.

Question

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 $X$.

Answer 1

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 $V$.