Proof of Extreme Value Theorem on a Compact Interval

Let

We first show that $E$ is nonempty. Since $f$ is continuous at $a$ in the sense of Continuity at a Point, there exists $\delta>0$ such that for every $x\in[a,b]$, if $|x-a|<\delta$, then

Thus $f(x)<f(a)+1$ whenever $x\in[a,\min\{a+\delta,b\}]$, so $f$ is bounded above on that initial subinterval. Therefore $E\ne\varnothing$.

The set $E$ is bounded above by $b$, so by Least Upper Bound Property of the Real Numbers there exists $c=\sup E$. We claim that $c=b$. Suppose instead that $c<b$. Because $f$ is continuous at $c$, there exists $\delta_c>0$ such that for every $x\in[a,b]$, if $|x-c|<\delta_c$, then

In particular, $f(x)<f(c)+1$ on $(c-\delta_c,c+\delta_c)\cap[a,b]$. Since $c=\sup E$, there exists $y\in E$ with $c-\delta_c/2<y\le c$. Because $y\in E$, the function $f$ is bounded above on $[a,y]$; say $f(x)\le M$ there. It follows that $f$ is bounded above on $[a,c+\delta_c/2]$ by $\max\{M,f(c)+1\}$. Hence $c+\delta_c/2\in E$, contradicting the fact that $c$ is an upper bound for $E$. Therefore $c=b$, and $f$ is bounded above on $[a,b]$.

Let

The set $S$ is nonempty and bounded above, so Least Upper Bound Property of the Real Numbers gives a real number $u=\sup S$ in the sense of Upper Bound and Least Upper Bound. For each integer $n\ge 1$, the number $u-1/n$ is not an upper bound for $S$, so there exists $x_n\in[a,b]$ such that

Since each $x_n$ lies in $[a,b]$, the sequence $(x_n)$ is bounded. By Bolzano-Weierstrass Theorem for Real Sequences, there is a subsequence $(x_{n_k})$ and a real number $x_{\max}$ such that $x_{n_k}\to x_{\max}$.

We claim that $x_{\max}\in[a,b]$. If $x_{\max}<a$, then with $\varepsilon=(a-x_{\max})/2>0$, the convergence $x_{n_k}\to x_{\max}$ implies $x_{n_k}<a$ for all sufficiently large $k$, contradicting $x_{n_k}\in[a,b]$. Similarly, $x_{\max}>b$ is impossible. Hence $x_{\max}\in[a,b]$.

Because $f$ is continuous at $x_{\max}$, it follows that $f(x_{n_k})\to f(x_{\max})$. On the other hand, from

we see that $f(x_{n_k})\to u$. Therefore

so $f$ attains its maximum on $[a,b]$.

Apply the same argument to the function $-f$. Since $-f$ is continuous at every point of $[a,b]$, there exists $x_{\min}\in[a,b]$ at which $-f$ attains its maximum. Equivalently, $f(x_{\min})\le f(x)$ for every $x\in[a,b]$. Thus $f$ attains both a minimum and a maximum on $[a,b]$.