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Let $X,Y$ be 2 topological spaces. $f\colon X\to Y$ be a mapping. It is known that

$f$ is continuous iff

  • $f[\bar{A}]\subseteq\overline{f[A]}$
  • $f^{-1}[\bar{B}]\supseteq\overline{f^{-1}[B]}$
  • $f^{-1}[\mathring{B}]\subseteq\overbrace{f^{-1}[B]}^\circ$

It is normally to ask that whether $f[\mathring{A}]\supseteq \overbrace{f[A]}^\circ$ can be added in this list?

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If ° is the interior operator, it should follow the set, e.g., $f[A^\circ]\supseteq f[A]^\circ$. –  Brian M. Scott Sep 26 '12 at 6:55
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Well, in many textbooks the little circle is placed above the set. This choice is painful with long expressions, though –  Siminore Sep 26 '12 at 7:08
    
@Siminore: Interesting. In $45$ years of reading them I’ve literally never seen that usage in a text in the U.S. –  Brian M. Scott Sep 26 '12 at 7:15
    
@BrianM.Scott What Siminore said. In fact, I thought the reason it sometimes follows the set on this site was people's laziness to find out the correct latex command to put it on top of the set. –  Rudy the Reindeer Sep 26 '12 at 7:48
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The function $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x) = 1$ for $x > 0$ and $f(x) = 0$ for $x \leq 0$ has the property $f[A]^\circ = \emptyset \subseteq f[A^\circ]$ for all subsets $A \subseteq \mathbb{R}$ but is not continuous. –  Matthias Klupsch Sep 26 '12 at 7:52

1 Answer 1

up vote 3 down vote accepted

Great question: I really had no idea how it was going to go. The answer is that $f[A^\circ]\supseteq f[A]^\circ$ is neither sufficient nor necessary for $f$ to be continuous.


Counterexample to Sufficiency

Generalizing from an example given in the comments, just send $X$ discontinuously to a set with empty interior. Then every subset of $X$ goes to a set with empty interior as well, and the condition holds trivially. It's not hard to make the function discontinuous: as long as $X$ isn't discrete and its image intersects two different open sets (e.g. is at least two points of a Haussdorff space), we can do it.


Counterexample to Necessity

Let $f:(X,\tau)\rightarrow (X,\sigma)$ where $X=\{a,b,c\}, \tau=\{X,\{a\},\{b,c\}\}, \sigma=\{X,\{a\},\{b\},\{a,b\}\}.$ Set $f(b)=f(c)=b, f(a)=a$. Then $f$ is continuous, since $f^{-1}(\{a,b\})=X,$ $f^{-1}(\{b\})=\{b,c\}$, and $f^{-1}(\{a\})=\{a\}$. But on the other hand $\{a,b\}$ has empty interior and is mapped to an open set $\{a,b\}$.

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Many thanks. It's quite a unexpected result. Besides, the first counterexample seems has a bug: I think the result is $f[A^{\circ}]\subseteq f[A]^{\circ}$. e.g. $A=[0,1)$. –  Popopo Sep 26 '12 at 11:10
    
You're right. I've edited. –  Kevin Carlson Sep 26 '12 at 11:26
    
Ok, that's all right. –  Popopo Sep 26 '12 at 11:28

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