why are closed disks considered open?

Question

I am doing a self-study through John Lee's Introduction to Topological Manifolds and am having trouble with Example 4.6(b) which says the following:

Example $4.6(b)$: Let $Y$ be the union of the two disjoint closed disks $\bar{B}_1(2,0)$ and $\bar{B}_1(-2,0)$ in $\mathbb{R}^2$. Each of the disks is open in $Y$, so the two disks disconnect $Y$.

In this notation $\bar{B}_1(2,0)$ means a closed disk centered at $(2,0)$ with radius $1$. I am confused because it doesn't seem like $\bar{B}_1(2,0)$ is open in $Y$. In particular, $Y$ is a subspace of $\mathbb{R}^2$. Further, from Proposition $3.5(b)$ from the previous chapter:

Proposition $3.5(b)$: Suppose $S$ is a subspace of the topological space $X$. If $U$ is a subset of $S$ that is either open or closed in $X$, then it is also open or closed in $S$, respectively.

Since in Example $4.6(b)$, $Y$ is closed in $\mathbb{R}^2$ and $\bar{B}_1(2,0)$ is closed in $\mathbb{R}^2$, it seems like we can conclude from Proposition $3.5(b)$ that $\bar{B}_1(2,0)$ is closed in $Y$.

How is $\bar{B}_1(2,0)$ open in $Y$? Thanks.

Answer 1

The sets $\bar B_1(2,0)$ and $\bar B_1(-2,0)$ are both closed in $Y$ by your argument. Hence their complements $Y\setminus \bar B_1(2,0)=\bar B_1(-2,0)$ and $Y\setminus \bar B_1(-2,0) = \bar B_1(2,0)$ are open in $Y$.

Sets are not doors... they can be open and closed.

Answer 2

$\overline{B}_1(2,0)=Y\cap B_{1+1/2}(2,0)$. Since the open sets of $Y$ are the intersections of open sets of $\mathbb{R}^2$ with $Y$ and $B_{1+1/2}(2,0)$ is open in $\mathbb{R}^2$, it follows that $\overline{B}_1(2,0)$ is open in $Y$.

It is also closed since $\overline{B}_1(2,0)=Y\setminus(Y\cap B_{1+1/2}(-2,0))$, which is the complement in $Y$ of an open set, $Y\cap B_{1+1/2}(-2,0)$ in $Y$.