a) Let $f : [a, b] → \mathbb{R}$ be a (not necessarily continuous) function with the property that, for every $x ∈ [a, b]$, there is a number $δ_x > 0$ for which $f$ is bounded on the neighborhood $Vδ_x (x)$ of $x$. Prove that the function f is bounded on the interval $[a, b]$.

b. Does the result of part a hold if the interval is $(a, b)$?

My proof for a: Let $k = [a, b]$. If $f(k)$ is not bounded, the for all $n \in \mathbb{N}$, there exist a sequence $\{y_n\}$ is a subset of $f(k)$ such that $|y_n| > n$. Let $\{x_n\}$ be a subset of $K$ satisfy $f(x_n) = y_n$. Then seq $\{x_n\}$ is bounded. So we extract convergent subsequences with $x_{n_j}$ converges to $x$ as $j$ goes to infinity. Since $K$ is closed, $x \in K$. By sequence criterion for continuity, $y_{n_j} = f(x_{n_j})$ converges to $f(x)$ . But $|y_{n_j}| > n_j \geq j$. This is a contradiction since unbounded sequences are divergences. Thus $f(k)$ is bounded.

Is this proof good ?

For part b, could someone give some hints how to proceed this proof?


1 Answer 1


For b take the function $f : (0,1) \rightarrow \mathbb{R}$, $f(x) = \frac{1}{x}$. The function is continuous in $(0,1)$. $f$ is bounded in every open nbd of $x \in (0,1)$ but it is unbounded in $(0,1)$.

Proof for a is good.

For proof a you can use property of compact set in $\mathbb{R}$.

A set $A \subset \mathbb{R}$ is said to be compact if each of its open cover has finite subcover. $[a,b]$ is a compact set in $\mathbb{R}$.

For any $x$ in $[a,b]$ we are getting an open nbd $V_{\delta}(x)$ of length $\delta$ where the function $f$ is bounded, i.e. $\exists$ $M_{\delta} > 0$ s.t. $f(x) < M_{\delta} \forall x \in V_{\delta}(x)$. Now $\{V_{\delta}(x) : x \in [a,b]\}$ is an open cover of $[a,b]$, having a finite subcover say $\{V_{\delta}(x_1), V_{\delta}(x_2) \dots V_{\delta}(x_n)\}$. $[a,b] = \cup_{k = 1}^n V_{\delta}(x_k)$. So get $M = \max\{M_1, M_2, \dots, M_n\}$. $f(x) < M$. Thus $f$ is bounded.

  • $\begingroup$ Is bounded by 1. Isn't it? $\endgroup$ Nov 22, 2013 at 1:34
  • $\begingroup$ The proof is OK, but uses continuity of $f$ ( while you claim initially that this is not a necessary condition). If $f$ is continuous, then certainly the image of a compact interval under $f$ is compact, so bounded. $\endgroup$
    – AnyAD
    Nov 22, 2013 at 1:46
  • $\begingroup$ However, as the question states, f is not necessarily continuous $\endgroup$ Nov 22, 2013 at 3:29
  • $\begingroup$ Is it then OK to use 'swquence criterion for continuity' of $f$ in the proof? $\endgroup$
    – AnyAD
    Nov 22, 2013 at 3:54
  • $\begingroup$ Do you have any idea to change the proof? $\endgroup$ Nov 22, 2013 at 4:44

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