# Proving open subset

• Apr 12th 2011, 10:39 AM
Borkborkmath
Proving open subset
This is coming from my introduction to topology class.

"Let (Y, d') be a subspace of the metric space (X, d). Prove that a subset $O' \subset Y$ is an open subset of (Y, d') iff there is an open subset O of (X, d) such that $O'=Y \cap O$. Prove that a subset $F'=Y \subset F$. For a point $a \in Y$, prove that a subset $N' \subset Y$ is a neighbohood of a iff there is a neighborhood N of a in (X, d) such that $N' = Y \cap N.$"

How and where do I start with this? The definition of a open subset?
A subset O of a metric space is said to be open if O is a neighborhood of each of its points.
• Apr 12th 2011, 12:43 PM
Deveno
how is the topology defined on Y? is d' the same metic as d? what is the subset F'?

let me give you an example of how a relative topology works, and why the open sets are different than in the larger space.

suppose X = R, the real numbers, and Y = [0,1], the unit interval. then the set (1/2,1] is not open in X, but it IS open in Y.

you see, if the definition of an open set was the same in Y as it was in X, the element 1 in Y would have NO neighborhoods at all!

and, by the very definition of a topology, the entire space is an open set. so Y = [0,1] is open, but [0,1] is certainly not open in X.

so the definition of a relative topology on a subspace Y of X is σ = {A∩Y: A is open in X}. if your subspace Y has the same metric, d,

then you should have either the statement about the N' or the statement about the O' as given.
• Apr 12th 2011, 01:39 PM
Borkborkmath
How does the metrics d' and d being different change the question? I feel a little lost in the topic of topology.

If the metrics are the same, then $O' = Y \cap O$ is a given? Then I would only have to show that $\leftarrow$ of the iff statement?
• Apr 12th 2011, 02:01 PM
Plato
Quote:

Originally Posted by Borkborkmath
How does the metrics d' and d being different change the question? I feel a little lost in the topic of topology. If the metrics are the same, then $O' = Y \cap O$ is a given? Then I would only have to show that $\leftarrow$ of the iff statement?

Actually when dealing with subspaces of metric spaces, $\left( {Y,d'} \right) \subseteq \left( {X,d} \right)$, it is generally understood that $d'$ is a restriction of $d$ to $Y$.

So if $p\in Y$ then $\mathcal{B}_Y \left( {p;\delta } \right) \subseteq \mathcal{B}_X \left( {p;\delta } \right)$. That is lemma to this proof.

The standard hint on how to proceed is to recall that any open set in a metric space is the union of basic balls, $\mathcal{B}_Y \left( {p;\delta } \right)$
• Apr 12th 2011, 02:30 PM
Borkborkmath
How do I show that $O' \subset Y$ is open then? Since it is a subset of open set Y, then O' could be closed since it could contain only one element.

I might be a little confused about what a metric space is too. Is it just a space where every element is defined as a distance from another element?
• Apr 12th 2011, 03:33 PM
Plato
Quote:

Originally Posted by Borkborkmath
I might be a little confused about what a metric space is too.

If that is really the case, it is pointless for you to attemp this question.
• Apr 12th 2011, 05:08 PM
ForumUserX
Quote:

Originally Posted by Borkborkmath
How do I show that $O' \subset Y$ is open then? Since it is a subset of open set Y, then O' could be closed since it could contain only one element.

I might be a little confused about what a metric space is too. Is it just a space where every element is defined as a distance from another element?

The element is not defined by the distance. The distance is defined between any two elements.

For example. In 3-space, a given point and distance define a sphere rather than one other point. But given 2 points, there is always a distance between them.
• Apr 12th 2011, 06:07 PM
Deveno
Quote:

Originally Posted by Plato
If that is really the case, it is pointless for you to attemp this question.

apparently, the poster is in a topology class, so it's a little late for that...

your earlier remark seems to indicate that the statement N' = N∩Y should be taken as given.

to the original poster: a topology is a set X, together with a distinguished collection of subsets of X, called open sets. there are basically just three rules such a collection must observe:

1) X and the null set are open.
2) the union of any two (and thus of infinitely many) open sets is open.
3) the intersection of any two (and thus of finitely many) open sets is open.

now, often, it is more convenient to deal with a smaller collection of open sets, called neighborhoods. these satisfy a weaker version of rule 3) if two neighborhoods contain a point x, then their intersection contains a neighborhood containing x. typically a neighborhood containing x is called a neighborhood of x, often written Nx, or N(x). then we recover a full topology by defining a set U to be open iff for every x in U there is a neighborhood N(x) contained in U.

if we have a metric function on a set, X, d, then we have some way of quantifying "how far apart" two points in the space are. in such a set, the neighborhoods are defined to be:

N_ε(x) = {y in X: d(x,y) < ε}, often referred to as "ε-balls" (although, depending on the metric, these may not even be round).

in your question, the point Plato is trying to make, is that if N' is an ε-ball in the relative topology for Y, N' = N∩Y, where N is an ε-ball in X.

the reasion it suffices to consider unions of ε-balls in a metric topology is this:

suppose we have two ε-balls N1 and N2 with N1∩N2 non-emtpy. for each point in N1∩N2, we can find some other ε-ball N(x), contained in the intersection.

if we take the union of all of these: U{N(x): x in N1∩N2}, this is clearly equal to N1∩N2. so N1∩N2 is open. clearly, all neighborhoods themselves are open,

so unions of neighborhoods are closed under intersection, and (trivially) unions of unions are. so unions of neighborhoods form a topology.
• Apr 12th 2011, 06:49 PM
Borkborkmath
Ty Deveno, we just covered what a topology is today.
We also covered a lot of things that you posted, but thank you anyhow.
I'll give this problem another go at with the new lecture.
• Apr 14th 2011, 11:10 PM
Borkborkmath
So, this question is really 3 smashed into one.
I want to attempt the first one, and I hope you guys can help me out.

Given that (Y, d') is a subspace of the metric space (X, d). Prove that subset O' $\subset$ Y is an open subset of (Y, d') iff there is an open subset O of (X,d) such that O; = Y $\cap$ O.

I'll start with $\Rightarrow$
We know that Y $\subset$X and open O' $\subset$Y
Let O = $\cup$(x $\in$O') B_x(x; $\delta$_x)
Then B_x(x; $\delta$_x) $\subset$O'
Showing O' $\subset$ (O $\cap$Y)
We know that O' $\subset$Y
if x $\in$O', we have x $\in$B_x(x, $\delta$_x) $\subset$O by construction
$\Rightarrow$ x $\in$O
O' $\subset$O
Thus O' $\subset$(O $\cap$Y)
Need to show containment both ways
O' $\subset$(o $\cap$Y) By construction of B_x(x; $\delta$_x) $\cap$Y $\subset$O'

*I need to show $\Leftarrow$ still :(*