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I'm having some trouble solving this problem:

Show that if $A$ is an open set in $(X, d_X)$, then a subset $G$ of $A$ is open in $(A, d_A)$ if and only if it is open in $(X, d_X)$.

It seems that if $G$ is a subset of $A$, and $G$ is open in $A$, then $G$ would be open in any set $X$ for which $A$ is a subset of $X$. Clearly this is incorrect, hence the question, so if someone could straighten out my logic, it would be much appreciated.

Edit: For reference, this question is section 1.3, question 8C from this document: https://www.uio.no/studier/emner/matnat/math/MAT2400/v11/Metric.pdf

Jendrik Stelzner
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Aaron
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  • Why do you think this is incorrect? – bsbb4 Sep 13 '18 at 08:52
  • @LukasKofler I think my reasoning is incorrect because it seems to contradict the premise of the question (i.e., the question implies A is proved from X but intuitively, it seems X being open for a subset G of A is a consequence of A being open because A is contained in X). Edit: But I don't disagree with the question's premise, if that's what you're asking. – Aaron Sep 13 '18 at 09:10
  • I feel like you‘re overthinking his. The question just tries to set out some basic facts; I think your original reasoning is correct. – bsbb4 Sep 13 '18 at 09:33
  • If you take $A\subseteq X$ a non-open subset, then $G=A$ is open in $A$ but not in $X$. For example $G=A={1}$ and $X=\mathbb R$. Another example would be $G=[0,1)$, $A=[0,2]$ and $X=\mathbb R$. Here $G$ is again open in $A$ but not in $X$. — So $A$ being open in $X$ is really necessary for the statement to be true. – Christoph Sep 13 '18 at 10:15

1 Answers1

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Assume $G$ is $A$-open.

Then $G=A\cap A'$ for some $X$-open $A'$.

Then $G$ is $X$-open since $A$ and $A'$ are both $X$-open.

Assume $G$ is $X$-open.

Then $G$ is $A$-open since $G=A\cap G$ for $G$ $X$-open.

FWE
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