About Theorem 17.4’s proof of Munkres’s Topology

general-topologyproof-explanation

Theorem 17.4
Let Y be a subspace of X; let A be a subset of Y; let $\overline{A}$ denote the
closure of A in X. Then the closure of A in Y equals $\overline{A}$ ∩ Y.
Proof. Let B denote the closure of A in Y. The set $\overline{A}$ is closed in X, so $\overline{A}$ ∩ Y is
closed in Y by Theorem 17.2. Since $\overline{A}$ ∩ Y contains A, and since by definition B equals
the intersection of all closed subsets of Y containing A, we must have B ⊂ ($\overline{A}$ ∩ Y ).
On the other hand, we know that B is closed in Y. Hence by Theorem 17.2,
B = C ∩ Y for some set C closed in X. Then C is a closed set of X containing A;
because $\overline{A}$ is the intersection of all such closed sets, we conclude that $\overline{A}$ ⊂ C. Then
($\overline{A}$ ∩ Y ) ⊂ (C ∩ Y ) = B.

Theorem 17.2
Let Y be a subspace of X. Then a set A is closed in Y if and only if
it equals the intersection of a closed set of X with Y.

Well, I am studying Topology from Munkres's Topology, and I am stuck here.

  1. What is the definition of the closure of A in a subspace Y ? In my intuition, the closure of A in a topological space X is "the smallest" closed set that exists in X and contains A.
  2. (What has really confused me) I couldn't understand "Then C is a closed set of X containing A" in the proof of Theorem 17.4. I know that C contains B, or C contains A ∩ Y, but I don't know anything about C ∩ (X\Y), for me, C ∩ (X\Y) could not contain A ∩ (X\Y).

Best Answer

Welcome to the site. See the MathJax Tutorial for some help with the formatting.

  1. What is the definition of the closure of $A$ in a subspace $Y$?

$Y$ forms a topological (sub)space. We can "forget" about $X$ and just ask the question: for a subset $A\subseteq Y$, what is the topological closure of $A$? This is all considered in the subspace topology of $Y$. There are many ways to define closure, the simplest being: $$\mathrm{cl}_Y(A):=\bigcap_{K:A\subseteq K\text{ and $K$ is closed in $Y$}}K$$Notice that $K$ being closed in $Y$ is not necessarily the same thing as $K$ being closed in $X$, so a priori the closure of $A$ in $Y$ is different to the closure of $A$ in $X$. In case I'm just rambling, let me reiterate that the definition is simply the same as the usual definition of closure, applied to the subspace topology of $Y$.

  1. [why does $C$ contain $A$?]

Your doubt is reasonable if $A$ is some 'arbitrary' subset of $X$. However, it says at the start that $A\subseteq Y$; $C$ contains $A\cap Y$ - as you say - but since $A\cap Y=A$, that just means: $C$ contains $A$.

Related Question