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In a proof of a question I found the following sentence:

"Since A⊂B, it follows that A∩V1 and A∩V2 are open in (A,d)."

Why does it hold?

I understand that if $x\in A\cap V_1\Rightarrow x\in A \wedge x\in V_1$. Now, $V_1$ is an open set, thus $\forall x\in V_1$ we can find an open ball $B(x,\epsilon)\subset V_1$.

Does it guarantee that $B(x,\epsilon)\subset A∩V_1$ (so that $A∩V_1$ to be open)?

I'm kind of confused. I think that i'm missing something. Can someone help me please?

Asaf Karagila
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Gr3gT
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    The claim is not that each $A\cap V_i$ is open (in $X$) but that $A\cap V_i$ is open in $A$ . Do you understand the difference? – Did Dec 08 '18 at 16:23
  • Thanks for the response. I understand that $A\cap V_i\subset A$. I'm confused only in the possible case that $A\subset V_i$ for $i=1$ or $i=2$. Does the claim hold because A is open and close in A? – Gr3gT Dec 08 '18 at 16:32
  • No, you are confused because you do not realize that being open in a subset of X is not synonymous to being open. Suggestion: check the definition of being open in a subset of X. – Did Dec 08 '18 at 16:33
  • Can you please send me a link or something? I found a definition about locally closed sets. Does something analogous hold for open sets too? I didn't find a definition of being open in a subset of X yet. – Gr3gT Dec 08 '18 at 16:55
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    It is by definition of the subset topology. – Nuntractatuses Amável Dec 08 '18 at 17:08
  • OK. Now i got it! Thank you. – Gr3gT Dec 08 '18 at 20:02

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