Problems on subspace topology and basis for a topology 
Need some help here. In Ex 2.35, I know that this is question of subspace topology and the basic definition is if there is $X$ which is a space with topology $T$ and $Y$ is a subset of $X$. $T$ induces a topology $TY$ on $Y$, called the subspace topology on Y, given by $TY:= \{V\text{ is a subset of }Y|\text{there exists }U\text{ in }T, \text{ s.t. }V =\text{intersection of } U\text{ and }Y\}$. Here, $Z$ is a subset of Y which means the subspace topology on Z, i.e. $Tz := \{V \text{ is a subset of }Z|\text{there exists }U \text{ in }TY,\text{ s.t. }V =\text{intersection of }U , Y \text{ and }Z\}$. Same thing for $Z$ as a subset of $X$, but how to show they will be same? 
Also, can you give some examples of a topology which has a basis? I am unclear as to the pairs which are going to form a basis. 
Any help will be greatly appreciated, thanks!!
 A: Noticing that all the open sets in X intersect with Z is the same as those open sets intersect with Y and Z. Since Z is a subset of Y, i.e. also the same as those open sets in X intersect with Z:
Let $T_X$ be the topology of X. The topology of $Y \subset X$ is $T_{XY} = \{O \cap Y, \forall O \in T_X\}$, the topology of $Z \subset X$ is $T_{XZ} = \{O \cap Z, \forall O \in T_X\}$, and the topology of $Z \subset Y$ is $T_{YZ} = \{O \cap Z, \forall O \in T_{XY}\} = \{O \cap Z \cap Y, \forall O \in T_{X}\} = \{O \cap Z, \forall O \in T_{X}\} = T_{XZ}$.
You could simply let $X = \mathbb{R}, Y = [0,2), Z = (0,1]$ and see how open sets act on those subsets.
A: Denote by $\mathcal{T}_Y$ the subspace topology on $Y$, inherited from $(X,\mathcal{T})$.
So $$\mathcal{T}_Y = \{O \cap Y: O \in \mathcal{T}\}$$
Likewise for $Z$ w.r.t. $X$:
$$\mathcal{T}_Z = \{O \cap Z: O \in \mathcal{T}\}$$
and we can also consider the subspace topology on $Z$ that it inherits from $\mathcal{T}_Y$:
$$\mathcal{T}' := \{O \cap Z: O \in \mathcal{T}_Y\}$$.
It's easy to see that $\mathcal{T}' = \mathcal{T}_Z$, by showing two inclusions:
let $U \in \mathcal{T}'$ then $U= O \cap Z$ with $O \in \mathcal{T}_Y$ so by definition $O = O' \cap Y$ with $O' \in \mathcal{T}$. But then
$$O' \cap Z = O' \cap (Y \cap Z)=(O' \cap Y) \cap Z = O \cap Z = U$$
so that $U \in \mathcal{T}_Z$. Note that $Z= Z \cap Y$ follows as $Z \subseteq Y$.
And if $U \in \mathcal{T}_Z$, write $U = O \cap Z$ with $O \in \mathcal{T}$. Then $O \cap Y \in \mathcal{T}_Y$ and so 
$$\mathcal{T}' \ni (O \cap Y) \cap Z = O\cap (Y \cap Z)=O \cap Z=U$$
and we've shown the other inclusion.
The remarks on "bases" are irrelevant for the stated problem. This is just a simple equality of certain families of subsets $\mathcal{T}'$ and $\mathcal{T}_Z$.
A: Z subset Y subset X.
The topology for Z subspace X, is
. . { U $\cap$ Z : U open within X }.  
The topology for Z subspace Y, is
. . { U $\cap$ Z : U open within Y }.
U is open within Y when exists V
. . open within X with U = V $\cap$ Y.
Thus U $\cap$ Z = V $\cap$ Y $\cap$ Z = V $\cap$ Z
. . is open within the subspace Z.  
Conversely, let U $\cap$ Z (U open within X)
. . be open within the subspace Z.
Thus U $\cap$ Y is open within Y.
Consequently, U $\cap$ Z = U $\cap$ Y $\cap$ Z
. . is open within the subspace Y of X.
