Given $f:\mathbb R\to[0,1]$ and a nonnegative sequence $\{u_n\}$, show that $\{f(u_n)\}$ has a convergent subsequence.

Let $f$ be a function from $\mathbb R$ to $[0,1]$. Prove that if $(u_n)$ is a sequence of non-negative real numbers, then there exists a subsequence $(u_{n_k})_{k\in\mathbb N}$ such that $(f(u_{n_k}))_{k\in\mathbb N}$ converges to a real number in $[0,1]$.

I'm not really sure how to go about starting this question. I wanted to first prove that Un is bounded above, then use the Bolzano-WeierstaB theorem to state that there has to exist a subsequence of Un which converges to R, and since the specific part of R [0,1] has already been defined, f(Unk) had to be in [0,1], but this sounds terrible and I'm very lost. Any help appreciated, thanks.

• The image of $f$ maps into a compact set, which is closed and bounded. Thus every sequence has a convergent subsequence. Mar 30, 2016 at 8:21

The assumption that $u_n\geqslant 0$ is irrelevant. Let $g_n=f(u_n)$. Since $\{g_n\}$ is a sequence in the compact space $[0,1]$, it has a subsequence $g_{n_k}$ with $\lim_{k\to\infty}g_{n_k}=g\in[0,1]$. Since $g_{n_k}=f(u_{n_k})$, the subsequence $u_{n_k}$ satisfies the claim.