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.

  • $\begingroup$ Do you have access to the compactness of $[0,1]$? Because this problem can be solved easily once we can utilize compactness. $\endgroup$ – Sangchul Lee Feb 23 at 6:29
  • $\begingroup$ I can use compactness. $\endgroup$ – user640290 Feb 23 at 6:30
  • 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.

  • $\begingroup$ Do you mean $0 < |x - c| < \delta$ in the first proof? @Sangchul Lee $\endgroup$ – user640290 Feb 23 at 6:46
  • $\begingroup$ @Diddysmash, Glad you pointed out my mistake :) Fixed it. $\endgroup$ – Sangchul Lee Feb 23 at 6:52
  • $\begingroup$ For the first proof, I'm not quite sure I understand your inequality. If $f$ is evaluated at some $c_i$, it could very well make $|f(c_i) - \ell|$ bigger than the $\varepsilon$ that worked for all the other $x \in U_{c_i}$. $\endgroup$ – user640290 Feb 26 at 2:24
  • $\begingroup$ @Diddysmash, That does not affect the boundedness of $f$ on each $U_{c}$. Indeed, an explicit bound of $f$ on each $U_c$ would be something like $$\sup_{U_c}|f| \leq \max\{|\ell|+2019, |f(c)|\}. $$ $\endgroup$ – Sangchul Lee Feb 26 at 4:49

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:

$\lim_{k \rightarrow \infty} x_{n_k} =X \in [a,b]$.


$\lim_{n \rightarrow \infty} y_n = L$ $ (< \infty)$,

hence bounded; a contradiction $(y_n >n, n \in \mathbb{Z^+})$.

  • $\begingroup$ The distinctness of the $x_n$ isn't clear from what you wrote, but you don't need it. $\endgroup$ – zhw. Feb 23 at 18:55

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