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I have a question that says:

Is there a continuous characteristic function on $\mathbb{R}$? If $A\subset \mathbb{R}$ show that $X_A$ is continuous at each point of $int(A)$. Are there any other points of continuity?

In this problem I have interpreted the characteristic function to be defined as:

$X_A: A \to \mathbb{R}$ where $X_A(x)=0$ if $x \notin A$ and $X_A(x)=1$ if $x \in A$

Here is my attempt:

$X_A(x)$ is continuous if given a closed set $E$ in the range of $X_A$ then $X_A^{-1}(E)$ is closed in $A$. E is closed in $R$ and in the range of the characteristic function only when $E= \emptyset$ $E=\{1\}$ and $E=\{0\}$ because singletons are closed in $\mathbb{R}$ since their compliment is the union of two disjoint open sets. $X_A(\{1\})= x\in A$ and $X_A(\{0\})=x\in \mathbb{R}-A$. But singletons in $ \mathbb{R} $ are closed so the inverse of a closed set is closed and $X_A$ is continuous.

Now for the second part of the question:

Suppose $x\in A^0$, the interior of $A$. Then $X_A(\{x\})$ is {1} which is closed in $\mathbb{R}$ hence $X_A$ is continuous.

Finally:

There are other points of continuity, those in $\mathbb{R}-A$ since the image of these points are closed subsets of $\mathbb{R}$ and their inverse image is closed.

I feel like I have flawlessly applied the definitions of continuity and given the exact answer the book is looking for!!! But somehow I suspect my feelings betray reality. Can you help me figure out how to fix this? In particular, I don't know if my using a singleton was the correct way to prove this, are singletons closed in R? I am pretty sure they are for the reason I gave.

Taylor Rendon
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  • Your proof appears to say that $X_A$ is continuous on $A$ and $\mathbb{R} \setminus A$, which is not correct (think $A=[0,1]$ for example). Instead, $X_A$ is indeed continuous on $\text{int}(A)$ and $\text{int}(\mathbb{R} \setminus A)$. – dxiv Oct 09 '16 at 21:44

1 Answers1

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Yes, the only such functions are in fact $\chi_{\mathbb{R}}$ and $\chi_{\emptyset}$, if I'm not mistaken. Any other characteristic function would have a jump at some point, which is necessarily a point of discontinuity on at least one side.

Matija Sreckovic
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  • I think then my proof is wrong completely. I thought $X_{{x}}(t)$ would be continuous at t = x. –  Oct 09 '16 at 21:40
  • By continuous characteristic function, do you mean a function continuous on the whole real line, $\mathbb{R}$ or continuous at some point $x \in \mathbb{R}$? – Matija Sreckovic Oct 09 '16 at 21:42
  • The question asks "Is there a continuous characteristic function on $\mathbb{R}$?" so I took that to mean that the function was continuous at some point in $\mathbb{R}$ –  Oct 09 '16 at 21:43
  • And, provided that $x$ is in the interior of $A$, the function $\chi_{A}$ will be continuous at $x$, because there will be a neighborhood of $x$ where the function is constant. – Matija Sreckovic Oct 09 '16 at 21:44
  • In that case, most functions indeed are, but $\chi_{{x}}$ from your example is, in fact, only discontinuous at $x$, because it's the only point where there's a jump. $\chi_{{x}}$ is continuous on $\mathbb{R} \setminus{x}$. Although I assume the author meant functions continuous on the whole real line, not just any point. – Matija Sreckovic Oct 09 '16 at 21:46
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    How could it ever be that the characteristic function is continuous across the whole line? It's range is {0,1} it seems like just by that alone it would never be continuous across the whole real line. –  Oct 09 '16 at 21:50
  • Let us continue this discussion in chat. – Matija Sreckovic Oct 09 '16 at 21:50