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Theorem:

Suppose $Y \subset X$. A subset $E \subset Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset G of X.

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I don't understand what's happening neither can I follow the proof.

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6  
For me, that's the defition of being "relatively open", what is your definition? –  martini Jul 12 '12 at 6:19
    
@martini I don't know that as yet. In fact, that's what I believe the theorem tries to teach (that it is necessary to describe the metric space within which the set is believed to be open unlike compactness since compactness is not relative) –  Inquest Jul 12 '12 at 6:21
    
@Inequest: So you are considering metric spaces ...? –  martini Jul 12 '12 at 6:24
    
@martini. I think so. (I don't know of any other possibilities) –  Inquest Jul 12 '12 at 6:26
    
I think it would be nice to include your definition of “relatively open” in the question! –  Dylan Moreland Jul 12 '12 at 6:29

2 Answers 2

up vote 2 down vote accepted

For metric spaces you can argue as follows:

First we note, that balls in $Y$ (which I will denote by $B^Y_\epsilon(y) = \{z \in Y \mid d(z,y) < \epsilon\}$ are intersections of balls in $X$ with $Y$, more concretely $B^X_\epsilon(y) \cap Y = B^Y_\epsilon(y)$: To prove this, first let $z \in B^Y_\epsilon(y)$, then $z \in Y$ and $d(z,y) < \epsilon$, hence $z \in B^X_\epsilon(y) \cap Y$. The other way round: If $z \in Y \cap B^X_\epsilon(y)$, then $z \in Y$ and $d(y,z) < \epsilon$, that is $z \in B^Y_\epsilon(y)$ by defition of a ball in $Y$. (If you want to draw a picture to see what is happening, you should imagine $y$ lying on the "boundary" of $Y$).

No to the proof of your theorem: First suppose that $E \subseteq Y$ is relatively open, that is for each $y \in E$ there is an $\epsilon_y > 0$ such that $B_{\epsilon_y}^Y(y) \subseteq E$. We want an open set in $X$ and only know something about balls, so we let $G = \bigcup_{y\in E} B^X_{\epsilon_y}(y)$. As a union of open balls in $X$, $G$ is clearly open in $X$ and (by our above lemma) \[ G \cap Y = \bigcup_{y \in E} B^X_{\epsilon_y}(y) \cap Y = \bigcup_{y\in E} B^Y_{\epsilon_y}(y) = E. \] Otherwise if $E = G \cap Y$ for some open $G \subseteq X$, given $y \in E$ we have an $\epsilon > 0$ with $B_\epsilon^X(y) \subseteq G$, hence $B^Y_\epsilon(y) = Y \cap B^X_\epsilon(y) \subseteq G \cap Y = E$. So $E$ is relatively open.

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A possibility for general topology: Let $(X,\tau)$ be a topological space and $Y \subset X$. We define the induced topology on $Y$ by $\tau'= \{ \Omega \cap Y | \Omega \in \tau \}$. So a subset of $Y$ is said to be relatively open in $Y$ if it is open in $Y$ for the induced topology.

Then, $Z \subset Y$ is relatively open in $Y$ iff there exists an open set $G \subset X$ such that $G \cap Y = Z$.

Indeed, if there exists such an open set $G$, $Z$ is relatively open by definition of the induced topology. If $Z$ is relatively open in $Y$ then for all $x \in Z$ there exists an open set $B_x$ of $X$ such that $x \in B_x \cap Y \subset Z$. Then, $Z= \left( \bigcup\limits_{x \in Z} B_x \right) \cap Y$.

Notice that the open sets of $Y$ are open in $X$ iff $Y$ is open. For example, $]0,1]$ is open in $[-1,1]$ but not open in $\mathbb{R}$.

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