Closed unit disk is connected

Question

A topological space $X$ is connected if it cannot be written as the disjoint union of 2 non-empty open subsets.

In trying to prove that the closed unit disk as a subset of $\mathbb{R}^2$ is connected I've run into an issue. First I supposed that the disk was disconnected and thus it is the disjoint union of 2 open subsets of $\mathbb{R}$. This then implies that the closed unit disk is open which is a contradiction.

However I realize that this argument falls apart if we examine the closed unit disk as its own topological space. In this space the closed unit disk is actually open, so we cannot create our contradiction.

My question is how would I prove this disk is connected in the subspace topology. More generally, if you try to prove that a subset is connected how do you deal with the changes to the open sets when we view it under the relative topology.

Answer 1

Hint: If a topological space is path-connected, then it is connected.

Answer 2

I assume here that you are using the metric topology as you are speaking of the closed unit disk.

Let us call $D$ the closed unit disk, and suppose that it not connected. There are 2 open subsets $A$ and $B$ verifying: $A\ne\emptyset, B\ne\emptyset, A\cap B=\emptyset$ and $A\cup B=D$. But as $A=D\setminus B$, and $B$ is open, $A$ is closed.

Considere the function $d_B: A\to\Bbb R$ defined by $d_B(x) = \inf(\{d(x, y), y\in B\})$. As $\{d(x, y), y\in B\}$ is a subset of $\Bbb R$ trivially minored by $0$, it has an inferior bound, so the function definition is valid.

Considere now the set $X = \{f_B(x), x\in A\}$. For the very same reason, $X$ has an inferior bound $d_0$. As the distance is continuous (by definition of the metric topology), and as $A$ is closed, $f^{-1}(d_0)\ne\emptyset$. So it contains at least one element $x_0$.

Considere now the function $d_{x_0}: B\to\Bbb R$, defined by $d_{x_0}(y)=\inf(\{d(x_0, y), y\in B\}$. By the same reasons that allowed to define $x_0$, we can find $y_0$ in B verifying $d(x_0, y_0) = d_0$.

If $d_0=0, x_0=y_0$ which is impossible because $A\cap B=\emptyset$.

So $d_0\ne 0$. Let us call $z_0$ the middle point of $(x_0, y_0)$. As the unit disk is convex, $z_0\in D$. So either $z_0\in A$, or $z_0\in B$.

If $z_0 \in A,d(z_0, y_0) = \frac{d_0} 2 <d_0$. Which is a contradiction

If $z_0 \in B, d(x_0, z_0) = \frac{d_0}2<d_0$. Which is a contradiction too.

The assumption $D$ is not connected leads to a contradiction, so $D$ is connected.

Answer 3

All you need to know are the following well-known facts:

1. The closed interval $[0,1]$ is connected.

2. If $X$ is connected and $Y$ is homeomorphic to $X$, then $Y$ is connected.

3. If $X = \bigcup_{\alpha \in A} X_\alpha$ with connected subspaces $X_\alpha \subset X$ such that $ \bigcap_{\alpha \in A} X_\alpha \ne \emptyset$, then $X$ is connected.

For each $x \in D := D^2 \setminus \{0\}$ let $L_x$ denote the line segment from $0$ to $x$. It is is homeomorphic to $[0,1]$, thus connected. We have $D^2 = \bigcup_{x \in D} L_x$ and $\bigcap_{x \in D} L_x = \{0\}$. By 3. $D^2$ is connected.

Note that this proof can be generalized to show that each star-shaped subset of $\mathbb R^n$ is connected.