E ⊆ Y ⊆ ( X , d )∣_(metric space) prove E is open in Y if and only if ∃ G (open set ) in X such that E = G ∩ Y .... (mathematical analysis (I))


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Question Number 206339 by mnjuly1970 last updated on 12/Apr/24

$E \subseteq Y \subseteq (X, d) ∣_{m e t r i c s p a c e}$ $p r o v e E i s o p e n i n Y i f a n d o n l y i f$ $\exists G (o p e n s e t) i n X s u c h t h a t$ $E = G \cap Y . . . . (m a t h e m a t i c a l a n a l y s i s (I))$
	Answered by Berbere last updated on 13/Apr/24
	$E=G∩Y G let a∈E;a∈G since G is open in X ∃ O∈X opene such that a∈O ⊂G O∩Y is open in Y since Y sub[space of (X,d) induit Topolopgiep (O∩Y)⊂(G∩Y)=E open in Y Suppose E is opened in Y E=∪_(e∈E) {e} ∀e ∃ O_e open in Y such That e∈O_e ⊂E O_e =O_e ′∩Y ,O′_e open in X G=∪_e O′_e opene in (X,d) ∀e O′_e ∩Y =O_e open in Y G ∩Y is open in Y ⇒G=∪_(e∈E) O′_e worcks$
	$E = G \cap Y G l e t a \in E; a \in G s i n c e G i s o p e n i n X$ $\exists O \in X o p e n e s u c h t h a t a \in O \subset G$ $O \cap Y i s o p e n i n Y s i n c e Y s u b [s p a c e o f (X, d) i n d u i t T o p o l o p g i e p$ $(O \cap Y) \subset (G \cap Y) = E o p e n i n Y$ $S u p p o s e E i s o p e n e d i n Y$ $E = \underset{e \in E}{\cup} {e}$ $\forall e \exists O_{e} o p e n i n Y s u c h T h a t e \in O_{e} \subset E$ $O_{e} = O_{e}^{'} \cap Y, O_{e}^{'} o p e n i n X$ $G = \cup_{e} O_{e}^{'} o p e n e i n (X, d)$ $\forall e O_{e}^{'} \cap Y = O_{e} o p e n i n Y$ $G \cap Y i s o p e n i n Y \Rightarrow G = \underset{e \in E}{\cup} O_{e}^{'} w o r c k s$
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