Can somebody please check if my proof is correct? And can it be done in a shorter way? (Using Bolzano Theorem)

A continuous function on a closed, bounded interval $[a,b$ attains a minimum at some point of $[a,b]$.


$f$ is continuous and bounded on $[a,b]$ such that $c,d\in[a,b]\subset \mathbb{R}$.

Let $\alpha =inf\left\{f(x):x\in [a,b]\right\}$.

$\implies \alpha+\frac 1n $ is not the greatest lower bound of $f$. Thus, there exists some mapping $x_n \rightarrow f(x_n) \ $ such that $f(x_n)$ lies in between its infimum $\alpha$ and $\alpha+\frac 1n$.

So, $\alpha \leq f(x_n) \leq \alpha+\frac 1n$.

Since $f$ is bounded on $[a,b]$, sequence $(a_n) =\left\{x_n\right\}_{n=1}^{n= \infty } \ \forall n\in\mathbb{N}$, is also bounded on $[a,b]$. Thus by Bolzano-Weierstrass theorem, $a_n$ contains at least one convergent subsequence in the interval $[a,b]$, say $a_{n_i}$ , such that $\lim_{i\to\infty}{ \ (a_{n_i})}=c$

Because f is continuous, we have: $$\lim_{i\to\infty}{ \ (a_{n_i})}=c\Rightarrow f(\lim_{i\to\infty}{ \ (x_{n_i}))}=f(c) \ for \ c\in [a,b]$$

By applying the squeeze theorem for functions, we note:

$$\alpha \leq f(x_n) \leq \alpha+\frac 1n \Rightarrow \lim_{n\to\infty}{ (\alpha)} = \lim_{n\to\infty}{f(x_n)}=\lim_{n\to\infty}{(\alpha+\frac 1n})=\alpha$$

So, $$\lim_{i\to\infty}{f(x_{n_i})=\lim_{n\to\infty}f(x_n)=\alpha}$$

This implies $\alpha :=inf\left\{f(x):x\in [a,b]\right\}=f(c)$

Thus the function f achieves a minimum value $f(c)$ for $c\in [a,b]$.

  • $\begingroup$ It is correct. Though, you don't need to apply squeeze Theorem. I don't see immediately how you could shorten this proof. Good Job. When you learn about compact sets, you'll learn ways to generalize the Mean Value Theorem to functions with very wild domains. $\endgroup$ – Melody Apr 7 at 1:34
  • $\begingroup$ The sequence $x_n$ is bounded not because of anything related to $f$ but because it's values are lying in $[a, b] $. Apart from that the proof is fine. $\endgroup$ – Paramanand Singh Apr 7 at 2:36

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