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Let $f:(X,d_X) \to (Y,d_Y) $ be continuous map where $X$ is compact. Prove that $f(X)$ is compact.

Proof using open cover:

Let $\{G_k | k\in J\}$ be an open cover of $f(X)$ so that

$f(X)\subseteq \cup_{k\in J}G_k \rightarrow X \subseteq f^{-1}(\cup_{k\in J}G_k ) = \cup_{k\in J} f^{-1}(G_k)$ where $f^{-1}(G_k)$ are open in $X$ because $f$ is continuous.

because $X\supseteq \cup_{k\in J} f^{-1}(G_k)$

we get, $X = \cup_{k\in J} f^{-1}(G_k)$$

Now $\{f^{-1}G_k |k\in J\}$ becomes open cover of $X$. As $X$ is compact, $\exists n\in N$ and $k_1\ldots k_n \in J$ such that

$X= \cup_{i=1}^{n}f^{-1}(G_k) \rightarrow f(X)=f(\cup_{i=1}^{n}f^{-1}(G_k)) = \cup_{i=1}^{n}f(f^{-1}(G_k)) \subseteq \cup_{i=1}^{n} G_k$

so we get a finite sub cover for $f(X)$. Hence it is compact.

Sequential argument:

Let $(y_n) \in f(X)$ be any sequence. $\exists (x_n)\in X$ such that $f(x_n)=y_n \;\forall n$.

As $X$ is compact, so $\exists (x_{n_k}$ subsequence of $(x_n)$ such that $(x_{n_k}) \to x_0 \in X$

By continuity of $f$, $(y_{n_k})=f(x_{n_k}) \to f(x_0) \in f(X)$

so $f(X)$ becomes compact.

Are both of these arguments correct?

chesslad
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    Yes. BTW this question has been answered in https://math.stackexchange.com/questions/874044/proof-that-the-continuous-image-of-a-compact-set-is-compact and https://math.stackexchange.com/questions/26514/proving-continuous-image-of-compact-sets-are-compact – user39082 Oct 24 '19 at 06:31
  • Note that the proof that sequential compactness implies compactness is a bit more involved. I am not convinced by what you have written. – Math1000 Oct 24 '19 at 08:54

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