# Image of Closed Real Interval is Bounded

## Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.

Then $f$ is bounded on $\closedint a b$.

## Proof

Suppose $f$ is not bounded on $\closedint a b$.

Then from the corollary to Limit of Sequence to Zero Distance Point, there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.

Since $\closedint a b$ is a closed interval, from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_n}$ which converges to some $\xi \in \closedint a b$.

Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.

But this contradicts our supposition that there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.

The result follows.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Proposition $5.1.1$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 9.11$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next):**continuous function**(vii)