Why does the notion of vanishing at infinity require local compactness|

Question

The definition I know of 'vanishing at infinity' for locally compact topological space is the following:

A function $f:X\to (Y,||\cdot||)$ on a locally compact space $X$ is said to vanish at infinity if for every $\epsilon>0$ the set $\{x\in X|\ ||f(x)||\geq\epsilon\}$ is compact.

However, I don't understand the necessity of the local compactness of $X$. To me it seems that if we skip that requirement, then on a non-locally-compact space, there will probably be only the null-function which vanishes at infinity, so the whole notion becomes a bit useless, but is still well-defined I think?

So can someone elaborate on why this is requirement is necessary?

Edit: I don't think it's true that only the null-function would vanish at infinity, but the function would have to be $0$ at non-locally-compact points.

Answer 1

Of course, the definition still makes sense for non locally compact spaces.

Also, non-null functions can be vanishing at infinity on non-locally compact spaces, for example on $X = \ell^2 (\Bbb{N}) \uplus [0,1]$, we can take $f = 1_{[0,1]}$.

But the author of the book/paper/whatever you are reading probably only needs this notion on locally compact spaces and thus does not bother to introduce the notion in greater generality. This is often the case for some definitions. Sometimes there might be a hidden deep reason, but in this case, I doubt there is.

Answer 2

Perhaps one reason for this restriction is that there are equivalent formulations which are only valid in the locally compact case. For example, $f$ vanishes at infinity if and only if the extension of $f$ to the one-point compactification $X \cup \{\infty\}$, obtained by setting $f(\infty)=0$, is continuous. This only makes sense when $X$ is locally compact because otherwise the one-point compactification does not exist.