Last time we didn't introduce the notion of a connected space. Let's do it now.

Sometimes, we can tell from a glance, whether a space is connected or not: the interval $[0,1]$ is connected, the set $\{1,2,3\}$ (with induced topology from $\R$) is not (there are gaps between points). But, we also know that topological space can be weird.

Ex: let $X = \{1,2,3,\cdots \}$ be a set, we declare the following subsets in $X$ are open:

• $\emptyset$ and $X$,
• $\{1,2,\cdots, n\}$, for some $n \geq 1$.

One can easily check that this collection of open sets are closed under arbitrary union and finite intersections. Hmm, is it connected? What does it mean when we ask this question?

Here is the general definition:

We say a metric space $X$ is connected, if $X$ cannot be written as disjoint union of two non-emtpy open subset. In other word, the only subsets in $X$ that is both open and closed are $X$ and $\emptyset$.

For example, $\Q$ is not connected, since $(-\infty, \sqrt{5})$ is both open and closed in $Q$. (why?)

For example, $X=\{1,2,3\}$ (with induced metric from $\R$) is not connected, since $\{1\}$ is both open and closed in $X$. (Discussion: Equip $X$ with the induced metric, can you show that $\{1\}$ is both open and closed? Equip $X$ with the induced topology, can you show that $\{1\}$ is both open and closed? )

For example, in that weird topology example above, $X$ is connected, since $X$ cannot be written as disjoint union of two open sets.

Theorem: a subset $E \In \R$ is connected, if and only if, for any $x,y \in E$, we have $[x,y] \In E$. Proof: suppose $E$ is connected, and $x,y \in E$, we need to show that $[x,y] \In E$. If not, say $z \in (x,y)$ is not in $E$, then let $E_1 = E \cap (-\infty,z)$, $E_2 = E \cap (z,+\infty)$, they are open subsets of $E$ (since they are open subsets of $\R$ intersecting $E$), they are non-empty (containing $x$ and $y$ respectively), and $E = E_1 \sqcup E_2$, therefore $E$ is not connected. Contradiction.

Suppose for any $x,y \in E$, we have $[x,y] \In E$, and suppose $E$ is not connected. Then $E$ can be written as disjoint union of two non-empty open sets $E = A \sqcup B$, where $A, B$ open in $E$. Pick $x \in A$ and $y \in B$, w.l.o.g, assuming $x<y$. By hypothesis, $[x,y] \In E$. Let $A' = A \cap [x,y]$ and $B' = B \cap [x,y]$. Then $[x,y] = A' \cup B'$, and $A', B'$ are open in $[x,y]$, and $x \in A'$, $y \in B'$. Let $z = \sup A'$. We discuss the following cases:

• if $z=x$, then $A' = \{x\}$, and $A'$ cannot be an open subset of $[x,y]$
• If $x< z < y$, we split into two case:
  • (a) if $z \in A'$, then $A'$ is not an open subset of $[x,y]$ at point $z$.
  • (b) if $z \in B'$, then $B'$ is not an open subset of $[x,y]$ at $z$.
• if $z = y$, then $z \in B'$, we claim that $B'$ is not open in $[x,y]$.

Thus, we get a contradiction.

Remark: Rudin uses a seeming different definition of connectedness, but it is equivalent. We say two subsets $A, B$ are 'separated', if $\bar A \cap B = \emptyset$ and $A \cap \bar B = \emptyset$. It is possible for two sets to be disjoint, but not separated, like $A = (0,1), B=[1,2)$.

Prop: A subset $E \In X$ is connected $\Leftrightarrow$ if $E$ cannot be written as the union of two non-empty separated sets.

Proof: $\Rightarrow$ Suppose $E \In X$ is connected, and $E = G \sqcup H$ with $G, H$ non-empty separated subsets in $X$. Then, $$ \bar G \cap E = \bar G \cap (G \sqcup H) = (\bar G \cap G) \sqcup (\bar G \cap H) = G \sqcup \emptyset = G $$, hence $G$ is closed in $E$. Similarly, $H$ is closed in $E$. That means the complement of $G$ in $E$, ie. $H$ is open, similarly, $G$ is open. Thus, $E = G \sqcup H$ writes $E$ as disjoint union of two non-emptyset open subsets of $E$, contradicting with $E$ being connected.

$\Leftarrow$: suppose $E$ is not connected, thus $E = G \sqcup H$ with $G, H$ open in $E$ (hence both $G, H$ are closed in $E$, since $G = E \RM H$, $H = E\RM G$). Let $G' \In X$ be a closed subset, such that $G' \cap E = G$, then $G' \cap H = \emptyset$. In particular, $\bar G \In G'$, hence $\bar G \cap H = \emptyset$. Similarly $\bar H \cap G=\emptyset$, thus $G,H$ are separated.