A subset $A\subseteq \mathbb{R}^n$ is compact if and only if the norm function defined on it is bounded and range of the function is closed.

Question

I don't know if the following result is true or not $ \colon $

A subset $A\subseteq \mathbb{R}^n$ is compact if and only if the norm function defined on it is bounded and range of the function is closed.

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My motive for such a statement is this and I have seen a couple of examples done this way. Hence I think such a statement should be true.
By Heine-Borel theorem we know the set is compact iff it is closed and bounded. Since the norm function is continuous and that if the range is closed then inverse image of the range is closed. Also if the range is bounded, then so is the set $A.$
Also I doubt that it is just "if" instead of "if and only if"
Any comments?

Answer 1

This is not true. For instance, with $n=1$, $A=[-1,1)$ is a counterexample. However, it is true with "only if" rather than "if and only if": if $A$ is compact, then the image of the norm function on $A$ is also compact since the norm function is continuous, so the norm function is bounded and has closed image.