I need to show that $f:[0,1] \rightarrow \mathbb{R}$ is bounded on $[a,b]$ if for all $c \in [a,b]$, $\lim\limits_{x \rightarrow c} f(x)$ exists.
I tried using the definition of limit, but I don't really know how to go about this completely. Any help hints would be great.
Proof using compactness: Let $c \in [0, 1]$ be arbitrary, and write $\ell$ for the limit of $f(x)$ as $x \to c$ in $[0, 1]$. Then there exists $\delta > 0$ such that $|f(x) - \ell| < 2019$ if $0 < |x-c| < \delta$. Then clearly $f$ is bounded on the open subset $U_c := (c-\delta, c+\delta) \cap [0, 1]$ of $[0, 1]$.
Now, since $[0, 1]$ is compact and $\{ U_c : c \in [0, 1]\}$ is an open cover of $[0, 1]$, there exists a finite sub-cover $U_{c_1}, \cdots, U_{c_n}$. Therefore
$$ \textstyle \sup_{[0,1]} |f| \leq \max \left\{ \sup_{U_{c_i}} |f| : i = 1, \cdots, n \right\} < \infty $$
and hence $f$ is bounded.
Proof using sequential compactness: Assume otherwise that $f$ is unbounded. Choose $(x_n)$ so that $|f(x_n)| \geq n$. Since $[0, 1]$ is sequentially compact, there exists a convergent subsequence $(x_{n_k})$. In doing so, we may assume that $(x_{n_k})$ are all different. If $c = \lim x_{n_k}$ denotes the limit of $(x_{n_k})$, then
$$ \lim_{x\to c} |f(x)| = \lim_{k\to\infty} |f(x_{n_k})| = \infty, $$
a contradiction.
You need some form of completeness of real numbers to prove this. Here is one approach via nested interval principle.
Suppose that $f$ is unbounded on $[0,1]$. Divide the interval $[0,1]$ into two equal subintervals via mid point and then $f$ must be unbounded on at least one of these subintervals. Repeat the procedure indefinitely to get a nested sequence of closed subintervals $I_n$ such that length of $I_n$ is $1/2^n$ and $f$ is unbounded on each $I_n$. By nested interval principle there is a unique $c\in[0,1]$ such that $c\in I_n\, \forall n\in\mathbb {N} $.
Since $\lim_{x\to c} f(x) $ exists the function $f$ is bounded in some open interval $I$ containing $c$ and we have an obvious contradiction as there is a value of $n$ for which $I_n\subseteq I$ (ask why and answer yourself).
You can also try other forms of completeness like supremum principle or Dedekind's theorem to prove this result.
Assume $f$ is not bounded.
For $n \in \mathbb{Z^+}$ there is a $x_n \in [a,b]$ s.t.
$y_n:= f(x_n) >n$.
The sequence $x_n$ is bounded.
Bolzano Weierstrass : Every bounded sequence of real numbers has a convergent subsequence $(x_{n_k})$.
$[a,b]$ is compact:
$\lim_{k \rightarrow \infty} x_{n_k} =X \in [a,b]$.
Given,
$\lim_{n \rightarrow \infty} y_n = L$ $ (< \infty)$,
hence bounded; a contradiction $(y_n >n, n \in \mathbb{Z^+})$.
