TITLE: Does this imply that the domain is bounded?
QUESTION [0 upvotes]: Suppose $B \subset \mathbb{R}^n$ and $f: B \to \mathbb{R}$ is a bounded function, does this imply that $B$ is also bounded? I was reading a proof about if $B$ is also closed, then $B$ is compact (by HB). So I am assuming this is true, but I am not sure why.
Why is $A$ compact then? I know $A$ is closed, but we don't know if it is bounded or not.

REPLY [1 votes]: The $f:B\to\mathbb R$ setting is unrelated to the question you ask about the proof in the book:

Why is A compact then?

Because A is a closed rectangle hence A is bounded and closed.
Later on, à propos another example, you state:

We know that $B_ϵ$ is closed, but nothing about being bounded.

Of course we know that $B_ϵ$ is bounded since $B_ϵ\subseteq A$ and $A$ is a rectangle.
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