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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we take X=R and Y=Z and E=N i.e. {1,2,3,4,5.........} then since for this case E is open in Y (as Y is itself an entire metric space) however there does not exist any open set G in X for this particular set. then how E is open in Y.

Pl clarify

share|cite|improve this question
1  
Why do you insist on extra notation? Just use $\mathbb{R,Z,N}$ instead of $X,Y,E$. Also what is "Pl clarify"? – Asaf Karagila Aug 25 '12 at 16:29
    
Also, why does $Y$ being an "entire metric space" imply that it is open in $E$? Any subset of a metric space is a metric space. $Y$ is open in $E$ in this case, but still... – ronno Aug 25 '12 at 19:52
    
These X, Y, and E are the letters from baby Rudin, though he does not specifically call them the reals, naturals, etc. This is think is a problem created to understand the theorem 2.30. Pl clarify:= Please Clarify. – in and out o' mind Oct 17 '15 at 8:36

Take $G = (0,\infty)$. $ $

share|cite|improve this answer

If you want to be fancy, let $$U=\bigcup_{n\in\Bbb Z^+}\left(n-\frac12,n+\frac12\right)\;;$$ then $U$ is an open subset of $\Bbb R$, and $U\cap\Bbb Z=\Bbb Z^+$. But there’s no need to go to so much trouble, as Sean’s answer shows. For that matter, you could observe that $\Bbb Z_{\le0}=\Bbb Z\setminus\Bbb Z^+$ is a closed set in $\Bbb R$, so $\Bbb R\setminus\Bbb Z_{\le0}$ is an open set in $\Bbb R$ whose intersection with $\Bbb Z$ is clearly $\Bbb Z^+$.

share|cite|improve this answer

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.