I'm reading loomis and this is left as an exercise but I'm not convinced I'm right. Can you please verify my proof and help me finish it? Also, what is the difference between compact and sequentially compact? Because up to this point the book talks about sequentially compact not compact, but the question is copy pasted from the book, so I'm a bit confused about terminology. Anyway here's the proof: $ \\ $
Let $f:X\rightarrow Y$ be continuous, $f(X)=Y$, and abuse notation by putting a metric $d$ on both $X$ and $Y$. Also $X$ is compact. I try to prove $Y$ is compact by showing any sequence $\{y_n\}\in Y $ has a convergent subsequence $\{y_{i(n)}\}\rightarrow y\in Y. $ Since $f$ is continuous, $\forall \tau>0, \forall n, \exists B_{\tau}(y_n)$ for which $\exists U_{n}\subset X $ that is open that is the preimage of $B_{\tau}(y_n)$. We can now define a sequence $\{x_n\}\in X$ by defining $x_n$ to be some arbitrary element in $ U_n$. Because $X$ is sequentially compact, $\exists \{x_{i(n)}\}\rightarrow x\in X \implies \forall \epsilon>0, \exists i(N)=K:\forall i(n)=k \geq i(N)=K, d(x_{i(n)},x)<\epsilon, i.e. x_{i(n)}\in B_{\epsilon}(x)$. Note that in this part of the proof I struggled to prove for big epsilon, but it still works for small ones I think. Please help for this to work for all epsilon not just small ones. Anwyay, $U_n$ is open so $\exists B_r:x\in B_r\subset U_n$, $B_r$ open by definition of $U_n$ being an open set. Thus we can easily pick $\epsilon<r$ so that we obtain $x_{i(n)}\subset B_{\epsilon}(x)\subset B_r(x)\subset U$. Now we get $f(x_{i(n)})\in f(B_{\epsilon}(x)) \subset f(U)=B_{\tau}(y_{i(n)})$. In particular because $f(x)\in f(B_{\epsilon}(x))\subset B_{\tau}(y_{i(n)})$ we have $f(x)\in B_{\tau}(y_{i(n)})$ and by definition $d(f(x),y_{i(n)})<\tau$. We can do this for all $\tau>0$ so that we get that the subsequence $\{y_{i(n)}\}\rightarrow f(x)$. This completes the proof. In addition to feeling insecured about the whole thing, I used $\epsilon<r$ which feels wrong, as for epsilon big, what forces x to be inside of U? I know it probably has to because the earlier statement holds for all epsilon, but if epsilon is large I don't think there's a way to show this. As a result I don't see why $f(x)$ would be in the ball centered at $y_n$, so really my proof is for epsilon less than r, so some ball contained in U. Also, as if this post wasn't bad enough, I used some metric here, could someone help to show this strictly topologically? Is it as simple as just dropping all the epsilon stuff and replacing balls with open neighborhood? Thanks. I'm looking forward to a reply.
Let $\{y_n\}$ be an arbitrary sequence in $Y$. Then since $f$ is surjective, each $y_n$ is of the form $y_n=f(x_n)$. Now, $\{x_n\}$ has a convergent subsequence, say $\{x_{a_n}\}$. Since $f$ is continuous, $\{f(x_{a_n})=y_{a_n}\}$ must be convergent. but this is a subsequence of $\{y_n\}$, so you are done.
P.S. Surjectivity of $f$ is absolutely necessary. Consider $[0,1]\subset\mathbb R$.