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.

Which of the following statements are true?

a. If $A$ is a dense subset of a topological space $X$, then $X \setminus A$ is nowhere dense in $X$.

b. If $A$ is a nowhere dense subset of a topological space $X$, then $X \setminus A$ is dense in $X$.

c. The set $\Bbb R$, identified with the x-axis in $\Bbb R^2$, is nowhere dense in $\Bbb R^2$.

I have tried to find out examples but could not succeed. Please Help.

share|improve this question
1  
If $A$ is a subspace of toplolgical space $X$, then what do you think $X \cap A$ is? It is A ifself, hence if A is dense in $X$ then $X \cap A$ is also dense in $X$ and if $A$ is nowhere dense then $X\cap A$ is also nowhere dense in $X$. You are only left with option C. First study some basics, then ask questions. –  Euclidean Sep 16 '12 at 3:33
    
that was not X∩A ,it was X-A(X difference A). i think you have mistaken somewhere.so edit the question properly –  poton Sep 16 '12 at 6:31
1  
poton, you wrote XnA, how is anbody supposed to figure out it is not $X \cap A$, and it is $A^c$. Post the questions correctly. And learn some latex, it's not that difficult. So edit your questions yourself, now. –  Euclidean Sep 16 '12 at 7:10
add comment

1 Answer

up vote 3 down vote accepted

HINTS:

(a) Think about $\Bbb Q$.

(b) This is true; if you understand the definitions of dense and nowhere dense, the proof is trivial.

(c) This also is true, and if you understand the definition of nowhere dense, the proof is trivial.

These are pretty small hints, because this problem really is extremely easy; if you’re really having trouble with it, you should probably talk to your instructor.

share|improve this answer
    
Regarding a: is it not true in general that if a set $A$ is dense in $X$ then $A^c$ is also dense in $X$? –  Matt N. Sep 19 '12 at 9:30
1  
@Matt: Absolutely not! What if $A=X$? –  Brian M. Scott Sep 19 '12 at 9:35
    
Oops. Do you also have a counter example if the subset is proper? (I have an excuse: I'm allowed to say non-sense since I got food poisoning 2 days ago and I'm still sick) –  Matt N. Sep 19 '12 at 9:36
    
Sure: $X=\Bbb R$, $A=\Bbb R\setminus\{0\}$. Or $A=\Bbb R\setminus\Bbb Z$. –  Brian M. Scott Sep 19 '12 at 9:36
    
Doh! Thank you. –  Matt N. Sep 19 '12 at 9:37
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.