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Apparently it is important that the support is defined as the closure of $f neq 0$. Because of that condition globalization is allowed as the exercise below indicates. However, I have no idea why it is so important that the support is defined as the closure of $f neq 0$?

The exercise:

Let (X, $mathcalT$) be a topological space, U $subset$ X open and $eta$ $in$ C(X)
, (C(X) is the space of continuous functions on X) supported in U. Then, for any continuous map $g:U rightarrow mathbbR $,

$(eta cdot g): X rightarrow mathbbR$,

$(eta cdot g)(x) = eta (x)g(x)$ if $x in U$ and

$(eta cdot g)(x) = 0$ if $x notin U$

is continous. Show that this statement fails if we only assume that $f neq 0 subset U$.

I have been able to show that the map $g : U rightarrow mathbbR $ is continuous. However, I still don't understand the importance of the closure and why the map otherwise isn't continuous.

Can anyone help me?

Imagine that you have a function $eta$ which satisfies $eta(x) neq 0$ for all $x in U$ while $eta(x) = 0$ for all $x in X setminus U$. If $X$ is connected and $U neq emptyset, X$ then such a function will satisfy $ , eta(x) neq 0 = U subseteq U$ but it won't satisfy
$$ overline , eta(x) neq 0 = overlineU subseteq U. $$

Take $g colon U rightarrow mathbbR$ to be the function $g(x) = frac1eta(x)$. Then $g$ is continuous on $U$ because $eta$ doesn't vanish on $U$ but "$eta cdot g$" is the characteristic function of $U$ so it's not continuous.

To see a concrete example, take $X = mathbbR, U = (-1,1)$ and
$$ eta(x) = begincases 1 - |x| & |x| < 1,\
0 & |x| geq 1. endcases$$
Then $eta cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = pm 1$.