# Why does the support must be closed?

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?

• You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $\eta \cdot g$ is continuous? – Chessanator Jan 12 at 18:43
• I meant that $(\eta \cdot g)$ is continuous yes! – HK4 Jan 12 at 19:46

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 $$x = \pm 1$$.