Closure and Interior of Subspace Let $A\subseteq Y \subseteq X$ be metric spaces. What does Cl$_X(A)$ or Cl$_Y(A)$  mean?
For example, let $X=[0,10],Y=[1,9]$ and $A=(4,6)$. 
Is $Cl_X(A)= Cl_Y(A)=[4,6]$? 
I understand that $Int(X)=(0,10), Int(Y)=(1,9), Cl(A)=[4,6]$, but the notation is a little confusing when finding the closure of a subspace and I'm unsure what it means. This question was a little helpful for reference:
Closure of subspace topology
 A: When $A$ is a subset of a topological space $X$ then the closure of $A$ is the smallest closed set of $X$ containing $A$. Analogously, the interior of $A$ is the largest open set of $X$ contained in $A$.
In cases where $A$ is a subset of more than one topological space of interest, as in the example $A\subseteq Y\subseteq X$ where $Y$ is a subspace of $X$, it is customary to note with respect to which topology the closure/interior is taken. i.e $Cl_X(A), Int_Y(A)$ etc.
This specification is important as even when $Y$ is a subspace of $X$ the closure of $A$ with respect to $Y$ may be different from the closure with respect to $X$. Same for interior. This happens because while the topologies of $Y$ and $X$ are intimately related, they are not the same topology, and hence have different closed and open subsets.
In your specific example, for instance, while $[1,9]$ is obviously not an open subset of $X$, it is an open subset of $Y$ (because it is the entire space). Therefore $Int_X(Y)$ is indeed $(1,9)$, but $Int_Y(Y)$ is $[1,9]$. Constructing an example where the closures differ is also possible, and not too difficult.
A: As already explained, $\operatorname{Cl}_Y$ and $\operatorname{Int}_Y$ are just the closure and interior with respect to the subspace topology $\{U \cap Y : U \text{ open in } X\}$ on $Y$.
You might find these formulae useful:

$$\operatorname{Cl}_Y (A) = \operatorname{Cl}_X(A) \cap Y$$
  $$\operatorname{Int}_Y (A) = \operatorname{Int}_X(A \cup( X \setminus Y)) \cap Y$$

Indeed,  $F$ is closed in $Y$ if and only if there exists $G$ closed in $X$ such that $F = G \cap Y$. Clearly $A \subseteq F \iff A\subseteq G$. 
$$\operatorname{Cl}_Y (A) = \bigcap_{A \subseteq F, F \text{ closed in } Y} F  = \bigcap_{A \subseteq G, G \text{ closed in } X} (G \cap Y) = \left(\bigcap_{A \subseteq G, G \text{ closed in } X} G\right)\cap Y = \operatorname{Cl}_X(A) \cap Y$$
Similarly, $U$ is open in $Y$ if and only if there exits $V$ open in $X$ such that $U = V \cap Y$. Clearly $U \subseteq A \iff V \cap Y \subseteq A \iff V \subseteq A \cup(X \setminus Y)$.
$$\operatorname{Int}_Y (A) = \bigcup_{U \subseteq A, U \text{ open in } Y} U  = \bigcup_{V \cap Y \subseteq A, V \text{ open in } X} (V \cap Y) = \left(\bigcup_{V \subseteq A \cup (X \setminus Y), V \text{ open in } X} U\right) \cap Y\\ = \operatorname{Int}_X(A \cup( X \setminus Y)) \cap Y$$
