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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 $X$.

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  • $\begingroup$ attempt added. @JoséCarlosSantos $\endgroup$ – J. Watson Oct 17 '18 at 6:53
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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 $V$.

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