Show that $\tilde{g}$ is continuous on $X.$

Question

Suppose that $X$ is a separable and completely metrizable space. Assume that $g:X \rightarrow \mathbb{R}$ is continuous.
Denote $A = \{ x \in X: g(x)<0 \}.$ Define $\tilde{g}:X \rightarrow \mathbb{R}$ given by $\tilde{g}(x)=g(x)$ if $x \in A$, $\tilde{g}(x) = 0$ if $x \not\in A.$

Question: Show that $\tilde{g}$ is continuous on $X.$

It suffices to check at 'boundary' of $A$, since $g$ and $0$ are continuous functions.

I do not know how to check at boundary of a set. Any hint would be appreciated.

There is a notion of checking continuity at a point for a topological space; although since both of your spaces are metrizable you could instead throw a metric on $X$ and work with the $\varepsilon-\delta$ version of continuity at a point. — Eric Stucky, Jun 02 '17 at 16:39

k.stm · Accepted Answer · 2017-06-02T17:02:01.407

2

If a map $f \colon X → Y$ is continuous on two closed subspaces $A, B ⊆ X$ covering $X$, i.e. $X = A ∪ B$, then it is continuous on all $X$.
So it suffices to check that $\tilde g$ is continuous on the closed sets $\operatorname{cls A}$ and $X \setminus A$ (as $A$ is open, being the preimage of an open set under a continuous function).
So it suffices to check that $\tilde g|_{\operatorname{cls} A} = g|_{\operatorname{cls} A}$, that is $g|_{∂A} = 0$.
Since $X$ is metrizable, so is $\operatorname{cls} A$. As $A ⊆ \operatorname{cls} A$ is dense, every point $x ∈ ∂A$ is the limit of a sequence in $A$. Note that $∂A ⊆ X \setminus A$, as $A$ is open in $X$, and so $g|_{∂A} ≥ 0$.

edited Jun 02 '17 at 17:02

answered Jun 02 '17 at 16:51

k.stm

19,187

By the way, I don’t think there’s a need for $X$ to be completely metrizable or separable. Being metrizable should be enough (unless I am missing some subtle point). – k.stm Jun 02 '17 at 17:01
What is the definition of $\partial A $? – Idonknow Jun 03 '17 at 00:22
@Idonknow It denotes the boundary of $A$. – k.stm Jun 03 '17 at 05:17

score 1 · Answer 2 · answered Jun 02 '17 at 16:51

1

Hint: $g$ is continuous, which intuitively means that is supposed to map close points close to each other. Formally this means that we obtain a $\delta$ such that [yadda yadda].

Notice that $\widetilde g$ maps close points either (1) equally close together or (2) closer together [if, say, g(point) is negative and g(other) is positive]. This gives us a strong guess as to a $\widetilde\delta$ that will work.

answered Jun 02 '17 at 16:51

Eric Stucky

13,018

(Note that this hint does not directly answer the question of how you'd check the boundary of a set. I talked about that in the comment above but I think the idea is a bit of a red herring.) – Eric Stucky Jun 02 '17 at 16:56

Show that $\tilde{g}$ is continuous on $X.$

2 Answers2