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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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    To me this looks like a definition rather than a theorem. Also, where is the proof that you can not follow?2012-07-12

2 Answers 2

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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 $\mathbb{R}$.