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From the Princeton book for the GRE Subject Test in Maths:


enter image description here


Me trying to prove proposition in red box (seems to be same as answer here, but there are no votes):

If $A$ is open in $S$ then $\exists \ O$ s.t.

  1. $O$ is open in $X$

  2. $$O \cap S = A$$

By assumption, $S$ is open in $X$.

By definition of topology, the intersection of two sets that are open in $X$ is open in $X$.

Hence, $A$ is open in $X$.

QED


What are some examples pertaining to the violation of the assumptions?

  1. If $A$ is not open in $S$, but $S$ is open in $X$, how can we not have $A$ is open in $X$?

  2. If $A$ is not open in $S$, but $S$ is open in $X$, how can we still have $A$ is open in $X$?

  3. If $A$ is open in $S$, but $S$ is not open in $X$, how can we not have $A$ is open in $X$? (already given in book)

  4. If $A$ is open in $S$, but $S$ is not open in $X$, how can we still have $A$ is open in $X$?


Notes:

  1. I'm not using Princeton as a replacement for textbooks or Schaum's. I'm using Princeton as a guide.

  2. I have done and plan to do more practice exams.

  3. I know topology may be in only at most 2 questions on an exam.

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If $S$ is open in $X$ and $A$ is not open in $S$, then $A$ is not open in $X$; for if it were then $A=A\cap S$ would be open in $S$.

If $A$ is open in $S$, but $S$ not open in $X$, it is of course possible that $A$ is open in $X$. Consider $X=\Bbb R$, $A=(0,1)$, $S=[0,1)$. But it is also possible that $A$ is not open in $X$. Consider $X=\Bbb R$, $S=\{0,1\}$, $A=\{0\}$.

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  • $\begingroup$ Thanks Hagen von Eitzen. 4 Oh right, what was I thinking. 1,2 I see...another question then, when we say '$A$ is not open in $S$', do we then deduce that $A \ne S$ since all sets are open in themselves? $\endgroup$ – BCLC Aug 3 '16 at 19:29
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If $X$ is a topological space $A \subset S \subset X$ and $S$ is open in $X$, then $A$ is open in $S$ if and only if $A$ is open in $X$.

It is easy to see that if $A$ is open in $X$, then $A = A \cap S$ is open in $S$.

Conversely, if $A$ is open in $S$, then by the definition of being open in $S$, there exists $G \subset X$, open in $X$, such that $A = G \cap S$. But the right hand side is the intersection of two open subsets of $X$, and hence is open in $X$.

I hope this answers point.

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