Question on step in proof of Weierstrass thereom (of max and min)

Let $$(X,d)$$ be a metric space and $$A\subset X$$ compact and not empty and $$f: A \to \mathbb{R}$$ a continuous function. Then f assumes its maximum and minimum.

Our proof

Since $$A$$ is compact, so is $$f(A) \subset \mathbb{R}$$ and therefore bounded and closed. Let $$y^* := \sup_{a \in A} f(a) \in \mathbb{R}$$. Now there exists a sequence $$(y_n)_{n \in \mathbb{N}} \subset f(A)$$ with $$y_n \to y^*$$.

And here's the problem But then there also exists a bounded sequence $$(x_n)_{n \in \mathbb{N}} \subset A$$ so that $$y_k = f(x_k)$$.

Why does this exist? If $$f$$ were invertible (it's continuous, so it would just have to be monotone as well), I would understand that the sequence $$x_k = f^{-1}(y_k)$$ is well defined for all $$k$$, but $$f$$ is not necessarily monotone.

The proof continues like this:

From the Bolzano-Weierstrass theorem we know that there exists a convergent subsequence $$(x_{n_k})_{k \in \mathbb{N}} \subset (x_n)_{n \in \mathbb{N}}$$ so that $$x_{n_k} \to x^*$$ for some $$x^* \in A$$.

Since $$f$$ is continuous, we have $$f(x_{n_k}) \to f(x^*)$$. On the other hand, we have $$f(x_{n_k}) = y_k \to y^*$$, and therefore $$y^* = f(x^*) \in f(A)$$. $$\square$$

By definition, $$f(A)=\{f(x):x\in A\}$$. Hence, if $$y\in f(A)$$, there is $$x\in A$$ with $$f(x)=y$$. No invertibility needed.
• Because it's a sequence in $A$ and any compact subset of a metric space is bounded. – Mars Plastic Mar 12 '19 at 17:20
For all $$k\in \mathbb{N}$$ you have that $$y_k\in f(A)$$ by construction. By definition of $$f(A)$$ there exists $$x_k\in A$$ such that $$y_k=f(x_k)$$.