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.