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Thm: Let $K$ be compact metric space and $f:K\rightarrow \mathbb{R}$ a continuous real-valued function. Then $f$ is bounded on $K$ and attains its infimum.

Since $K$ is compact and $f$ continuous then $f(K)$ is a compact subset of $\mathbb{R}$ and therefore is bounded by the Heine-Borel theorem.

Since $f$ is bounded, it is bounded from below and the infimum $m$ on $K$ is finite.

By the definition of the infimum, for each $n\in\mathbb{N}$ there is an $x_n\in K$ such that $$m\leq f(x_n) < m+\frac{1}{n}\hspace{10mm} (1)$$ This inequality implies that $$\lim_{n\rightarrow\infty}f(x_n) = m\hspace{10mm} (2)$$

I know by the definition of the $\inf f$ that $m\leq f(x)$ for all $x$.

How does inequality $(1)$ happen? How can you have function values greater than or equal to this $m$ but also strictly less than $m+\frac{1}{n}$?

Also how does this inequality imply $(2)$? If you take $n\rightarrow\infty$ doesn't the inequality become $m\leq f(x_n) < m$?

This may seem very trivial but any help and comments would be greatly appreciated. Thank you.

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    $\begingroup$ You started in the middle of an argument, without context. What are the hypotheses on $f$ and $m$? $\endgroup$ – Jonas Meyer Oct 1 '13 at 4:45
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I am assuming from context that $m$ is the infimum.

Suppose that there were an $n$ for which there weren't such an $x_n$. Then for every $x$, we'd have $f(x) \ge m + \frac{1}{n}$; but this will contradict the definition of infimum. In particular, the infimum would need to be at least as large as $m + \frac{1}{n}$.

For your second question, note that inequalities can get weaker in the limit. So $<$ will frequently become $\le$. One way to see this is to consider a sequence $x_n = \frac{1}{n}$; then each term satisfies $x_n > 0$, but that doesn't hold in the limit.

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  • $\begingroup$ Great thank you. This definitely helped! $\endgroup$ – RDizzl3 Oct 1 '13 at 4:53

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