The following is the beginning of the proof in the lecture notes of the Multi-Dimensional Extreme Value Theorem:
$$
\begin{array}{l}{\text { Proposition } 4.20 \text { Let } X \text { be a closed bounded set in } \mathbb{R}^{m}, \text { and let } f: X \rightarrow \mathbb{R}^{n}} \\ {\text { be a continuous function mapping } X \text { into } \mathbb{R}^{n} . \text { Then there exists a point }\mathbf{w}} \\ {\text { of } X \text { such that }|f(\mathbf{x})| \leq|f(\mathbf{w})| \text { for all } \mathbf{x} \in X .} \\ \\{\text { Proof: Let } g: X \rightarrow \mathbb{R} \text { be defined such that }} \\ {\qquad g(\mathbf{x})=\frac{1}{1+|f(\mathbf{x})|}} \\ {\text { for all } \mathbf{x} \in X . \text { Now the function mapping each } \mathbf{x} \in X \text { to }|f(\mathbf{x})| \text { is continuous }} \\ {\text { (see Lemma 4.6) and quotients of continuous functions are continuous where}\\ \text{ they are defined (see Lemma }4.5) .}\end{array}
$$
$$
\begin{array}{l}{ \text { It follows that the function } g: X \rightarrow \mathbb{R} \text { is }} {\text { continuous. }} {\text { Let }} { m=\inf \{g(\mathbf{x}): \mathbf{x} \in X\} .} \\ {\text { Then there exists an infinite sequence } \mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \ldots \text { in } X \text { such that }} \\ {\qquad g\left(\mathbf{x}_{j}\right)<m+\frac{1}{j} \\ \text{for all positive integers }j.}\end{array}
$$
How do we know that such a sequence exists?
Best Answer
This follows definition of infimum (you do not need continuity of $g$ for this part). Suppose that there exists some $j\in\mathbb{N}$ such that for all $x\in X$ $$|g(x)| \geq m + \frac{1}{j}$$ but then for any $\epsilon >0$, in particular take $\epsilon$ small enough so that $\frac{1}{j} - \epsilon >0$, we have $$m = \inf_{x\in X} |g(x)| \geq m + \frac{1}{j} - \epsilon > m$$ a contradiction. Thus there must be some $x_j$ such that $$|g(x_j)| < m + \frac{1}{j}$$ Then you use continuity to conclude that if $x_j \rightarrow x^*\in X$ (by compactness), then $$m \leq \lim_{j\rightarrow \infty} |g(x_j)| = g(x^*) \leq \lim_{j\rightarrow \infty}(m+\frac{1}{j}) = m$$ therefore $|g(x^*)| = m$.