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Proof

(related to Proposition: Continuous Real Functions on Closed Intervals Take Maximum and Minimum Values within these Intervals)

• By hypothesis $[a,b]$ is a closed real interval and $f:[a,b]\to\mathbb R$ is continuous.
• We take $A$ as the supremum of the image set $\{(f(x)\mid x\in[a,b]\}.$ Please note that $A$ is either a real number or $A=\infty$ if the image set is unbounded.
• We have to show that $A$ is a real number (which is equivalent to $f(x)$ is bounded on $[x,y]$).
  • The definition of a supremum of the image set is equlivalent to there is a convergent real sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in[a,b]$ $\lim_{n\to\infty} f(x_n)=A.$
  • Since $[a,b]$ is a bounded subset of real numbers, the real sequence $(x_n)_{n\in\mathbb N}$ with $x_n\in[a,b]$ is a bounded sequence.
  • Since every bounded real sequence has a convergent subsequence, so does $(x_n)_{n\in\mathbb N}.$
  • In other words, there is a subsequence $(x_{n_k})_{k\in\mathbb N}$ of $(x_n)_{n\in\mathbb N}$ such that $$\lim_{k\to\infty} x_{n_k}=\alpha$$ for some $\alpha\in[a,b].$
  • Since $f$ is continuous, we have $$\lim_{k\to\infty} f(x_{n_k})=f(\alpha).$$
  • Therefore, $A=f(\alpha).$ In particular $A$ is a real number and it is the maximum of the image set.

Note: The proof for the existence of a minimum can be formulated analogously, replacing $f$ by $-f$, $\sup$ by the infimum $\inf,$ and $\max$ by the minimum $\min.$

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983