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Claim:Let $S\subset T\subset X$ where $X$ is a metric space. If $T$ is compact in $X$ then $S$ is also compact in $X$.

Proof:Given that $T$ is compact in $X$ then any open cover of T, there is a finite open subcover, denote it as $\left \{V_i \right \}_{i=1}^{N}$. Since $S\subset T\subset \left \{V_i \right \}_{i=1}^{N}$ so $\left \{V_i \right \}_{i=1}^{N}$ also covers $S$ and hence $S$ is compact in X

Edited: I see why this is false but in general, why every closed subset of a compact set is compact?

  • 9
    You need to prove it for any open cover of $S$, not just the covers of $S$ that also cover $T$2012-10-13
  • 1
    O i see which part goes wrong, since $S\subset T$ there may exist a subcover of $S$ which may not cover $T$ and that subcover may not finite at all.2012-10-13
  • 4
    No, there may exist a *cover*of $S$ that does not cover $T$2012-10-13
  • 0
    Please do your best to use informative titles. As it is right now, it is impossible to figure out what the question is about from the title and tags.2012-10-13

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