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I am having some trouble dealing with the following

Let $\Omega \subseteq \mathbb{R}^m$ be closed and bounded, and $f:\Omega \to \mathbb{R}^n$ continuous in $\Omega$. Without using Heine-Borel theorem, show that $f(\Omega)$ is closed.

Can I find some sequence $(f(\mathbf{x}_k))$ such that $f(\mathbf{x}_k) \to f(\mathbf{x})$, with $f(\mathbf{x}) \in f(\Omega)$? Is it the right approach? If so, how do I build it or show it exists?

Thanks!

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Yeah, precisely. You would want to look at an arbitrary, convergent sequence $f(x_n) \subset f(\Omega)$ and show that this sequence has its limit in $f(\Omega)$, i.e. $$\lim_{n\rightarrow \infty} f(x_n)=f(y)$$ for some $y \in \Omega$. If you have the Bolzano-Weierstass theorem at your disposal, since $\Omega$ is closed and bounded, you know that the corresponding sequence of preimages, $x_n$, has a convergent subsequence $x_{n_k} \in \Omega$, i.e. $$\lim_{k\rightarrow \infty} x_{n_k} = y \in \Omega$$ Now, we're almost there. We just need to answer two questions:

1.) What does the subsequence $f(x_{n_k})$ tend to in $f(\Omega)$? (hint: it has something to do with the continuity of $f$).

2.) What does this tell us about the initial sequence $f(x_n$)? (Which we supposed to be convergent)

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    Thank you! Can I assume there is a sequence $(x_n) \in \Omega$ without exposing it? – 2017-02-26
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    No problem. The subsequence $x_{n_k}$ you don't need to expose, but the sequence $x_n$ is derived from the sequence $f(x_n)$ as such: $x_n = f^{-1}(f(x_n))$. It's just the preimages of the values of $f$ in the sequence. – 2017-02-26
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You are almost there. In fact, you need some subsequence $(f(\mathbf{x}_{n_k}))$.

Let ${\bf y}\in \overline{f(\Omega)}$. There exists a sequence $({\bf x}_n)_n$ in $\Omega$ such that $f({\bf x}_n) \to {\bf y}$ as $n \to +\infty$. Consider the projection $p_i:\Bbb R^m \to \Bbb R$ defined as $p_i({\bf x}_n) = p_i((x_{n,1},x_{n,2},\dots,x_{n,m})) = x_{n,i} \,\forall {\bf x}_n \in \Bbb R^m \, \forall i \in \{1,\dots,m\}$. In other words, $p_i$ is the projection of a vector ${\bf x} \in \Bbb R^m$ into its $i$-th component.

We start from $i = 1$ (the first component): $(p_1({\bf x}_n))_n = (x_{n,1})_n$ is bounded in $\Bbb R$, so by Bolzano-Weierstrass Theorem, $(x_{n,1})_n$ has a subsequence $(x^{(1)}_{n,1})_n$ such that $x^{(1)}_{n,1} \to x_1$ as $n \to +\infty$. Now we have a subsequence $({\bf x}^{(1)}_n)_n$ of $({\bf x}_n)_n$ so that its first component converges.

Repeat this for $i = 2$: $(p_2({\bf x}^{(1)}_n))_n = (x^{(1)}_{n,2})_n \to x_2$ since it's bounded in $\Bbb R$. Now we have a subsequence $({\bf x}^{(2)}_n)_n$ of $({\bf x}^{(1)}_n)_n$ so that its first two components converge.

Inductively, we have a subsequence $({\bf x}^{(k-1)}_n)_n$ with first $k-1$ components converge. From this, we extract a subsequence $({\bf x}^{(k)}_n)_n$ of $({\bf x}^{(k-1)}_n)_n$ so that the first $k$ components converge.

Since $i \in \{1,\dots,m\}$, this process will end and we have a subsequence $({\bf x}^{(m)}_n)_n$ of $({\bf x}_n)_n$ so that each of its components converges. Therefore, ${\bf x}^{(m)}_n \to {\bf x} := (x_1,\dots,x_m)$ as $n \to +\infty$. Since $\Omega$ is closed, ${\bf x} \in \Omega$, so $f({\bf x}) \in f(\Omega)$. Note that

  1. $f({\bf x}^{(m)}_n) \to {\bf y}$ as $({\bf x}^{(m)}_n)_n$ is a subsequence of $({\bf x}_n)_n$.
  2. $f({\bf x}^{(m)}_n) \to f({\bf x})$ as $f$ is continuous and ${\bf x}^{(m)}_n \to {\bf x}$ by construction.

Since the limit is unique, we conclude that $f({\bf x}) = {\bf y}$. Hence, for any ${\bf y}\in \overline{f(\Omega)}$, ${\bf y}\in f(\Omega)$, so $f(\Omega)$ is closed.

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    Thank you. Bringing in $\overline{f(\Omega)}$ helps! – 2017-02-26