Could we maybe show it also as follows?
We suppose that $f(A)$ is not bounded.
There is also a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$. Or do we not get that when $f(A)$ is not bounded?
Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ it follows that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$, or not?
$\{f^{-1}(x_n)\}_{n=1}^{\infty}$ is therefore, bounded, right?
From Bolzano-Weierstrass there is a convergent subsequence, let $\{f^{-1}(x_{n_k})\}_{k=1}^{\infty}$ with $f^{-1}(x_{n_k})\rightarrow y\in \mathbb{R}$.
Since $f$ is continuous, we have that $f\left (f^{-1}(x_{n_k})\right )\rightarrow f(y)\in \mathbb{R}$.
That means that $x_{n_k}\rightarrow f(y)$.
But how could we get a contradiction?
Last edited by mathmari; January 5th, 2017 at 05:13.
It could be that there is no sequence in $f(A)$ whose limit is infinity, given that $f(A)$ is unbounded. But what is true for sure is that that is a sequence whose limit is either $+\infty$ or $-\infty$. We may WLOG assume that the limit is $+\infty$.
Note that $f^{-1}(x_n)$ is a subset of of $\mathbf R$. So $\{f^{-1}(x_n)\}$ is not a sequence in $\mathbf R$.
So the way to go is by choose $y_n\in f^{-1}(x_n)$ for each $n$. Then you get a sequence $\{y_n\}$ in $\mathbf R$.
Now from Bolzano-Weierstrass there is a convergent subsequence $\{y_{n_k}\}_{k=1}^{\infty}$ with $y_{n_k}\rightarrow y\in \mathbb{R}$.
Since $f$ is continuous, we have that $x_n=f(y_n)\rightarrow f(y)\in \mathbb{R}$.
That means that $x_{n_k}\rightarrow f(y)$. The contradiction is that since $x_{n_k}$ is a subsequence of $\{x_n\}$, we also have $x_{n_k}\to \infty$.