TITLE: Subspace Topology of a subset
QUESTION [5 upvotes]: Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.
First some notational specifications: Let us define $\tau$ as the topology of $X$
$\tau_{Y}=\{U \cap Y; U \in \tau \}$; $\tau_{A}^Y=\{V \cap A; V \in \tau_{Y} \}$ and $\tau_{A}^X=\{U \cap A; U \in \tau \}$
Let us show that $\tau_{A}^Y=\tau_{A}^X$
$\subset$: Let $V \cap A$ be an element of $\tau_{A}^Y$ such that $V \in \tau_{Y}$
Then, there exists $U\in \tau$ such that $V=U\cap Y$
$V\cap A=(U\cap Y)\cap A=U\cap (Y\cap A)= U\cap A$
Hence: $V\cap A \in \tau_{A}^X$ and so: $\tau_{A}^Y \subset \tau_{A}^X$
$\supset$: Let $U \cap A \in \tau_{A}^X$
$U\in \tau \iff U \text{ open in } X$ 
But $U \cap Y$ open in $Y \iff U \cap Y \in \tau_{Y}$
Hence $(U \cap Y) \cap A \in \tau_{A}^Y$ since $U \cap Y \in \tau_{Y}$
Therefore $U \cap A \in \tau_{A}^Y$
Hence: $\tau_{A}^X \subset \tau_{A}^Y$
Therefore: $\tau_{A}^Y=\tau_{A}^X$

REPLY [4 votes]: Let $\tau_{X}$ and $\tau_{Y}$ be the topologies of A as subspace of X and Y.
If $U \in \tau_{X} \Rightarrow \exists B  $ open in X such that $U = B \cap A = (B\cap Y) \cap A$ and then $U \in \tau_{Y}$.
If $U \in \tau_{Y} \Rightarrow \exists B  $ open in Y such that $U = B \cap A \Rightarrow \exists C $ open in X such that $ B = C \cap Y \Rightarrow U = (C \cap Y)\cap A = C \cap ( Y \cap A) = C \cap A.$
Then $\tau_{X} =\tau_{Y}$.