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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

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  • $\begingroup$ I feel my answer (and others) could be made clearer if we knew how familiar you are with the concept of a subset topology. $\endgroup$ – Bar Alon Mar 25 '18 at 22:40
  • $\begingroup$ I haven't learned what a subspace topology was, I think I meant to just say subspace in the title. The problem I was having was resolved once I realised that, for example, $[-1,1]$ is the largest open subset of the space $[-1,1]$. Thank you for your help! $\endgroup$ – Elijah Moore Mar 27 '18 at 11:42
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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.

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  • $\begingroup$ Thanks for the response. How can $Int_Y(Y)=[1,9]$ if the interior of a set is always open? $\endgroup$ – Elijah Moore Mar 25 '18 at 2:18
  • $\begingroup$ @ElijahMoore $[1,9]$ is open in $Y = [1,9]$. $\endgroup$ – Henno Brandsma Mar 25 '18 at 14:24
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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 $U \subseteq A \iff V \cap Y \subseteq A \iff V \subseteq A \cup(X \setminus Y)$.

$$\operatorname{Int}_Y (A) = \bigcup_{U \subseteq A, U \text{ open in } Y} U = \bigcup_{V \cap Y \subseteq A, V \text{ open in } X} (V \cap Y) = \left(\bigcup_{V \subseteq A \cup (X \setminus Y), V \text{ open in } X} U\right) \cap Y\\ = \operatorname{Int}_X(A \cup( X \setminus Y)) \cap Y$$

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