Image of Closed Real Interval is Bounded

Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a .. b}\right]$.

Then $f$ is bounded on $\left[{a .. b}\right]$.

Proof

Suppose $f$ is not bounded on $\left[{a . . b}\right]$.

Then from the corollary to Limit of Sequence to Zero Distance Point, there exists a sequence $\left \langle {x_n} \right \rangle$ in $\left[{a .. b}\right]$ such that $\left|{f \left({x_n}\right)}\right| \to +\infty$ as $n \to \infty$.

Since $\left[{a . . b}\right]$ is a closed interval, from Convergent Subsequence in Closed Interval, $\left \langle {x_n} \right \rangle$ has a subsequence $\left \langle {x_{n_r}} \right \rangle$ which converges to some $\xi \in \left[{a .. b}\right]$.

Because $f$ is continuous on $\left[{a . . b}\right]$, it follows from Limit of Image of Sequence that $f \left({x_{n_r}}\right) \to f \left({\xi}\right)$ as $r \to \infty$.

But this contradicts our supposition that there exists a sequence $\left \langle {x_n} \right \rangle$ in $\left[{a .. b}\right]$ such that $\left|{f \left({x_n}\right)}\right| \to +\infty$ as $n \to \infty$.

The result follows.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 9.11$