I'm looking for guidance to improve my proof. It's a proof I gave on an exam and it's all over the place because I couldn't figure out what to focus on.
Let $f:(0,1]\rightarrow\mathbb{R}$ be a continuous function and let $x_n$ be a sequence in the domain $(0,1]$. Suppose that the sequence $f(x_n)$ is unbounded. Show that $x_n$ has a subsequence that converges to $0$.
Here's what I understand:
- The sequence is unbounded, so no $M$ exists such that $-M<f(x_n)<M$.
- $f$ is continuous, so it limits to something at $x=0$, i.e. $\lim_{x\rightarrow0}f(x)=\infty, -\infty, \text{ or }c$, where $c$ is some real number.
- Every bounded sequence has a convergent subsequence.
Here's my pitiful attempt at a proof:
Let $f:(0,1]\rightarrow\mathbb{R}$ be continuous.
Let $\{x_n\}$ be a sequence in $(0,1]$.
Let $\{f(x_n)\}$ be unbounded.
By definition of boundedness and continuity, $f(1)$ must exist, so $\{f(x_n)\}$ is bounded above.
But $\{f(x_n)\}$ is not bounded below.
Because ${x_n}$ is bounded, it has a convergent subsequence.
Coupled with the fact that $f$ is defined on $(0,1]$, the subsequence converges to $0$.
$f$ is not said to be increasing, so the existence of $f(1)$ cannot deduce that $\{f(x_n)\}$ is bounded above.
Since $\{f(x_n)\}$ is unbounded, we can wlog assume that $\{f(x_n)\}$ is not bounded above, i.e., there is a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $f(x_{n_k})\to+\infty$ as $k\to+\infty$. The boundedness of $\{x_{n_k}\}$ implies that there is a subsequence $\{y_l\}$ of $\{x_{n_k}\}$ such that $y_l\to a$ for some $a\in[0,1]$. If $a\neq 0$, the continuity of $f$ gives that $f(y_l)\to f(a)$, contradicting to the assumption that $f(y_{n_k})\to+\infty$. Hence $y_l\to0$, as desired.