# Is this function bounded on $[a,b]$?

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.

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

$$\lim_{x\to c} |f(x)| = \lim_{k\to\infty} |f(x_{n_k})| = \infty,$$

• Do you mean $0 < |x - c| < \delta$ in the first proof? @Sangchul Lee – user640290 Feb 23 at 6:46
• @Diddysmash, Glad you pointed out my mistake :) Fixed it. – Sangchul Lee Feb 23 at 6:52
• 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 $x \in U_{c_i}$. – user640290 Feb 26 at 2:24
• @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 $$\sup_{U_c}|f| \leq \max\{|\ell|+2019, |f(c)|\}.$$ – 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:

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

Given,

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

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

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