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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, $\mathcal{T}$) 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 \mathbb{R} $,

$(\eta \cdot g): X \rightarrow \mathbb{R}$,

$(\eta \cdot g)(x) = \eta (x)g(x)$ 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 \mathbb{R} $ 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?

LK4
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1 Answers1

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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 $\{ x \in X \, | \, \eta(x) \neq 0 \} = U \subseteq U$ but it won't satisfy $$ \overline{ \{ x \in X \, | \, \eta(x) \neq 0 \} }= \overline{U} \subseteq U. $$

Take $g \colon U \rightarrow \mathbb{R}$ to be the function $g(x) = \frac{1}{\eta(x)}$. 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 $X = \mathbb{R}, U = (-1,1)$ and $$ \eta(x) = \begin{cases} 1 - |x| & |x| < 1,\\ 0 & |x| \geq 1. \end{cases}$$ Then $\eta \cdot g$ is the characteristic function of $(-1, 1)$ which is discontinuous at $x = \pm 1$.

levap
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