1
$\begingroup$

Let $E$ be a topological space and $A\subset E$ how to find a necessary and sufficient condition for the continuity of the function $\chi_A: E\rightarrow\mathbb{R}$ where $$ \chi_A(x)= \begin{cases} 1, ~x\in A\\ 0,~x\not\in A \end{cases} $$

If i suppose that $\chi_A$ is continuous then $$\forall\varepsilon>0, \exists V\in \mathcal{V}_x, \chi_A(V)\subset ]\chi_{A}(x)-\varepsilon, \chi_A(x)+\varepsilon[$$ or $$\forall\varepsilon>0, \chi_A^{-1}(]\chi_{A}(x)-\varepsilon, \chi_A(x)+\varepsilon[)\in \mathcal{V}_x $$

How to find a condition on $A$?

Thank you

  • 1
    "The current answer(s) are out-of-date and require revision given recent changes" Rather, the recent changes are quite inadequate and should never have been made after two detailed answers were posted (and explained in comments). Sorry but this revision and this bounty are just not how the site is supposed to function. As you have been made aware in the comments, "you do not respect our work to give the answers". – 2017-03-09

3 Answers 3

3

The inverse image (by a continuous function) of a closed set is closed, so $A$ should be closed as the inverse image of a singleton $\{1\}$. Similarly, the complement of $A$ should be closed as an inverse image of $\{0\}$. Then $A$ should be a clopen set. If $A\in\{\emptyset,E\}$, the characteristic function is constant, so it is continuous. This is the only possibility in a connected space. Now let's consider a proper clopen set $A$ in a disconnected space. Could you continue with the check whether $\chi_A$ is continuous?

  • 0
    @Vrouvrou, the question is quite OK, I don't know why you want to delete it. – 2017-02-26
  • 0
    There's nothing wrong with that. We can mark this question as a duplicate if you find a link to the other one. At best it will help people in the future with the same problem to find answers easier. – 2017-02-26
  • 3
    @Vrouvrou, ... and by deletion of your question you do not respect our work to give the answers. – 2017-02-26
  • 2
    @Vrouvrou, I am sorry, I don't intend to delete my answer. – 2017-02-26
  • 0
    really i don't understand your answer why you begin by the inverse image of a closed set is closed/.....???? – 2017-03-03
  • 0
    @Vrouvrou, there are many methods for solving problems. I have chosen this one. My conclusion is correct, so also the method is correct. – 2017-03-03
  • 0
    i don't understand if $A$ is clopen how to prove that the function is continuous ? please @szw1710 – 2017-03-03
  • 0
    @szw1710 can i say that $x\in A$ as $A$ is open then $\forall \varepsilon>0, \exists V=A\in \mathcal{V}_x, \chi_A(x)\subset ]1-\varepsilon,1+\varepsilon[$ so $\chi_A$ is continuous on $x\in A$ and by the same way whwn $x\notin A$ ? – 2017-03-03
2

Necessary and sufficient condition for $\chi_A$ to be continuous is that $A$ is both open and closed.

Necessity. If $\chi_A$ is continuous, then $\chi_A^{-1}(-\infty,1/2)=X\setminus A$, $\chi_A^{-1}(1/2, \infty)=A$ are open.

Sufficiency. Assume that $A$ and $X\setminus A$ are both open, and $U\subset\mathbb R$ open. Then
$$ \chi_A^{-1}(U)=\left\{ \begin{array}{ccc} X &\text{if}& 0,1\in U,\\ A&\text{if}& 1\in U \,\&\, 0\not\in U\\ X\setminus A&\text{if}& 0\in U \,\&\, 1\not\in U\\ \varnothing&\text{if}& 0,1\not\in U. \end{array} \right. $$ and hence $\chi_A$ is continuous.

1

Let's work locally. Take $x_0 \in E$. If $\chi_A$ is continuous at $x_0 \in E$, for every $\epsilon>0$ there is a neighbourhood $V$ of $x_0$ such that $x \in V$ implies $\chi_A(x) = \chi_A(x_0)$. So for all points in $A$ (for which $\chi_A(x_0)=1$, we find a neighbourhood of it which is contained in $A$, and so $A$ is open. Similarly for $A^c$ replacing $1$ by $0$ we find that $A$ is closed.

So $\chi_A$ continuous implies $A$ clopen. Conversely, if $A$ is clopen, take $(x_\alpha)$ a net converging to $x$. If $x \in A$, then $x_\alpha \in A$ for large $\alpha$ and so $\chi_A(x_\alpha) \to 1$. If $x \not\in A$, since $A^c$ is open, a similar argument shows that $\chi_A(x_\alpha)\to 0$. So $\chi_A$ is continuous.

  • 1
    I'm happy our conclusions are the same. :-) – 2017-02-26
  • 0
    What is a net converging to x ? – 2017-02-26
  • 1
    Your way is more elegant, but I used another approach exactly to double check haha – 2017-02-26
  • 0
    @Vrouvrou A net is sort of a generalization of a sequence, where you can index the elements with things other than $\Bbb N$. Since we don't know if your space is first-countable, I couldn't use sequences. – 2017-02-26
  • 0
    we don't know what is this so i can answer like this – 2017-02-26
  • 0
    For now you can read "sequence" instead of "net" and "$n$" instead of "$\alpha$" and little to no harm will be done (at least here). If you haven't seen it in class, better to avoid talking about nets. Consider this answer an incentive to look it up. Folland's book is a nice reference. – 2017-02-26
  • 0
    @IvoTerek have you an other method to prove the seconde implication ? – 2017-03-02
  • 0
    You can use szw's idea and work with neighbourhoods. Take $x \in A$. Let $0 <\epsilon <1$. By continuity you get a neighbourhood $V$ of $x$ such that $y\in V$ implies $\chi_A(y)=\chi_A(x) $. This means that $V\subseteq A$, and so $A$ is open. Repeat for $E\setminus A$ and fill any details. – 2017-03-02
  • 0
    i don't understand your answer, my definition of continuity is $$\forall \varepsilon>0, \exists V\in \mathcal{V}_{x}, \chi_A(x)\subset]\chi_A(x)-\varepsilon, \chi_A(x)+\varepsilon[$$ – 2017-03-03
  • 0
    You are forgetting three basic facts: (1) a set is open if and only if every point has an open neighbourhood contained in the set , (2) if $|\chi_A(x_1) = \chi_A(x_2)| <1$, then $\chi_A(x_1)=\chi_A(x_2)$, because $\chi_A$ only assumes two values ($0$ and $1$) whose distance is less than $1$, and (3) continuity is a hypothesis, so you get to choose $\epsilon$ - in particular, any $\epsilon <1$. – 2017-03-03
  • 0
    i don't understand the chronology of the answer – 2017-03-03
  • 0
    @IvoTerek i start with the continuity , after that how we think to find that $A$ is open and closed? – 2017-03-03
  • 0
    We prove that $A$ is open by checking that for all $x\in A$ there is an open neighbourhood $V$ of $x$ such that $V\subseteq A$. If this seems too alien for you, think that $E$ is a metric space and $V$ is an open ball centered in $x$. The continuity of $\chi_A$ gives one such $V$ for each $\epsilon <1$. The exact same argument proves that $E\setminus A$ is open. – 2017-03-04