Proving open subsets

[borkborkmath]

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

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.

 

[pka]

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

Tell us exactly what you know about metric spaces.
If you are not fully conversant with the basic properties of metric spaces, this question is beyond your grasp.

 

[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' \(\displaystyle \subset\) Y is an open subset of (Y, d') iff there is an open subset O of (X,d) such that O; = Y \(\displaystyle \cap\) O.

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

*I need to show \(\displaystyle \Leftarrow\) still *

 

[pka]

Here is the whole basis of this proof. It is the very reason you were asked to do this proof.
It shows you a fundamental relationship.

Fundamental principle: In any metric space an open set is the union of fundamental balls.

If (Y,d’) is a subspace of (X,d) then any fundamental ball in Y is a subset of a fundamental ball in X. So one of the directions is trivial.

Now suppose that \(\displaystyle \mathcal{B}_X(p;\delta)\) is a ball in \(\displaystyle X\) then clearly \(\displaystyle \mathcal{B}_Y(p;\delta)=Y\cap \mathcal{B}_X(p;\delta)\).
There you have it.

 

[borkborkmath]

wait..thats it?
Putting your line and my other stuff together answers the first question?