# Open sets in a metric space

• Mar 18th 2011, 04:16 PM
Jame
Open sets in a metric space
The question is:

Let $\displaystyle (X,d)$ be a metric space, and $\displaystyle Y$ be a subset of $\displaystyle X$.

If $\displaystyle G \subset X$ is open, then $\displaystyle G \cap Y$ is open in the metric space $\displaystyle (Y,d)$.

Pf:

$\displaystyle Y \subset X$ implies $\displaystyle (Y,d)$ is a metric space $\displaystyle \rightarrow Y$ is open.
Therefore $\displaystyle Y^{C}$is closed

$\displaystyle G$ is open, thus $\displaystyle G^{C}$ is closed.

So $\displaystyle B = (G^{C} \cup Y^{C})$ is closed, since it is the finite union of closed sets.

Therefore $\displaystyle B^{C} = G \cap Y$ is open.

I seem to follow all the definitions and theorems correct. Is this ok?

There is also a second part to this question.

Conversely, if $\displaystyle E_1 \subset Y$ is open in $\displaystyle (Y,d)$, $\displaystyle \exists$ an open set $\displaystyle E \subset X$ s.t $\displaystyle E_1 = E \cap Y$

I don't get this question. Don't we already have an open set that works in the hypothesis? E = E_1?

Any help is greatly appreciated. Thank you.

-Jame
• Mar 18th 2011, 04:25 PM
Plato
Quote:

Originally Posted by Jame
Let $\displaystyle (X,d)$ be a metric space, and $\displaystyle Y$ be a subset of $\displaystyle X$.
If $\displaystyle G \subset X$ is open, then $\displaystyle G \cap Y$ is open in the metric space $\displaystyle (Y,d)$.

I do not understand your confusion!
How are open sets in a sub-space related to open sets in the super-space?
• Mar 18th 2011, 05:57 PM
Jame
Oh right. The are not necessarily related. Something could be open in X but not in a subset of X. Guess I have to start over... and use the definition of open and not try to be tricky with complements.
• Mar 18th 2011, 07:43 PM
Jame
Actuallly I have a question.

This is how my book defines open

For a metric space $\displaystyle (X,d)$ a set $\displaystyle G \subset X$ is open if $\displaystyle \forall x$ in $\displaystyle G$ $\displaystyle \exists$ an $\displaystyle \epsilon > 0$ s.t $\displaystyle B(x; \epsilon) \subset G.$

This defiinition doesnt seem to depend on X. Does it matter the space we are in?
• Mar 19th 2011, 02:44 AM
Plato
Quote:

Originally Posted by Jame
This is how my book defines open
For a metric space $\displaystyle (X,d)$ a set $\displaystyle G \subset X$ is open if $\displaystyle \forall x$ in $\displaystyle G$ $\displaystyle \exists$ an $\displaystyle \epsilon > 0$ s.t $\displaystyle B(x; \epsilon) \subset G.$
This defiinition doesnt seem to depend on X. Does it matter the space we are in?

Actually it does depend upon $\displaystyle X~\&~d$.
That is why we use the notation $\displaystyle (D,d)$.
A metric space is determined by a pair, a set of points and a metric on the set.
Recall that a ball is defined as $\displaystyle \mathcal{B}(x;\delta)=\{y\in X: d(x,y)<\delta\}.$
Balls are the basic open sets of a metric space.
• Mar 20th 2011, 11:58 AM
Jame
Oh yes!. Thanks so much. I was honestly didn't catch that. I was looking at this definition and was wondering "whats the difference between being open in X and in Y?" Thank you for the clarification.