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I need th proving this theorem:

Let $(X, d)$ is metric space and let $(X_1, d_1)$ is its subspace. The set $G_1\subset X_1$ is open set in $X_1$ if and only if there is an open set $G$ in $X$ such that $G\cap X_1=G_1$. The set $F_1\subset X_1$ is closed set in $X_1$ if and only if there is an closed set $F$ in $X$ such that $F\cap X_1=F_1$.

Prove: First part of the theorem I've proved i.e. (The set $G_1\subset X_1$ is open set in $X_1$ if and only if there is an open set $G$ in $X$ such that $G\cap X_1=G_1$.)

But I do not know how to prove the second part of the theorem (The set $F_1\subset X_1$ is closed set in $X_1$ if and only if there is an closed set $F$ in $X$ such that $F\cap X_1=F_1$).

Please if anyone can help me to prove the second part of the theorem, thank you, for your help and your attention

Madrit Zhaku
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So $\;F_1\subset X\;$ is closed in $\;X_1\;$ iff $\;X\setminus F_1\;$ is open iff there exists open $\;G\subset X\;$ s.t.

$$(X\setminus F_1)\cap G\;\;\text{is open}\iff X\setminus\left((X\setminus F_1)\cap G\right)\stackrel{\text{de Morgan}}=\left(F_1\right)\cup\left(X\setminus G\right)\;\;\text{is closed}$$

(since $\;X\setminus(X\setminus F_1)=F_1\;$ ) , and we're done since trivially

$$F_1=F_1\cap\left[\left(F_1\right)\cup\left(X\setminus G\right)\right] $$

DonAntonio
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  • writing that should be used the De Morgan`s formula, whether this is a complete solution, thank you – Madrit Zhaku Nov 07 '13 at 19:01
  • @MadritZhaku, after reading your comment three times, I still have no idea what you tried to convey to me... – DonAntonio Nov 07 '13 at 19:04
  • Sorry, just wanted to know if your solution is complete and should be continued but I'm sure your solution is a complete and thank you very much – Madrit Zhaku Nov 07 '13 at 19:07
  • Oh, ok @MadritZhaku, but that's a weird question since then it seems to be you didn't understand completely the solution (as then you'd see it is complete)... – DonAntonio Nov 07 '13 at 19:08