Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $(X,d)$ and $(X',d')$ are metric spaces and $f:X\rightarrow X'$ is continuous.

(a) If $A\subseteq X$ and $x_o$ is an isolated point of $A$, then $f(x_o)$ is an isolated point of $f(A)$.

Attempt: So an isolated point of $A$ means that $\exists r>0$ s.t. $B_r(a)\cap A=\{a\}$. Since if $f$ is continuous an open set $V\subset X'$ means that $f^{-1}(V)$ is open in $X$, can I use this fact somehow to show that the map preserves the isolated point?

(a) If $A\subseteq X$, $x_o\in A$ and $f(x_o)$ is an isolated point of $f(A)$ then $x_o$ is an isolated point of $A$.

Attempt: Same deal as above -- I'm not sure if I'm thinking of the right theorem in proving this.

share|improve this question
1  
Are you sure about the first part? What if I do something silly like mapping the discrete set $\mathbf Z_{\geq 0} \to \mathbf R$ by sending $0 \mapsto 0$, $n \mapsto 1/n$ otherwise? –  Dylan Moreland Feb 6 '12 at 4:03
    
Or, less silly, $f:\mathbb R\to\mathbb R$ by $x\to x^2$ and $A=\{-1\}\cup[0,2]$? –  Henning Makholm Feb 6 '12 at 4:13
3  
The second part is false too, consider the map from $\mathbb R \to \mathbb Z_{\geq 0} $ by sending everything to $0$. –  azarel Feb 6 '12 at 4:15
1  
I should probably have mentioned that the question had a "true or false" component -- assuming that the statements were true was indeed quite naive! –  Emir Feb 6 '12 at 5:22
2  
There really isn't any harm in telling us everything you know about the problem :) –  Dylan Moreland Feb 6 '12 at 16:06
show 1 more comment

1 Answer 1

up vote 4 down vote accepted

Both parts are often false. For the first part, see the comment of Dylan Moreland, which even gives a counterexample that is injective.

For the second part, consider constant functions. All you can say is that if $f(x_0)$ is isolated in $f(A)$, then there is a relatively open subset of $A$ containing $x_0$ where $f$ is constant.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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