4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ You were given that $g$ was continuous as a hypothesis. Did you mean that you proved that $\eta \cdot g$ is continuous? $\endgroup$ – Chessanator Jan 12 at 18:43
  • $\begingroup$ I meant that $(\eta \cdot g)$ is continuous yes! $\endgroup$ – HK4 Jan 12 at 19:46
5
$\begingroup$

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 $x = \pm 1$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.