0
$\begingroup$

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

  • 1
    Why do you insist on extra notation? Just use $\mathbb{R,Z,N}$ instead of $X,Y,E$. Also what is "Pl clarify"?2012-08-25
  • 0
    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...2012-08-25
  • 0
    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.2015-10-17

2 Answers 2