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Math4201 Topology I (Lecture 22)
Connectedness of topological spaces
Connectedness
Definition of separation
A separate of a topological space X is a pair of disjoint nonempty open subsets U,V\subset X such that X=U\cup V. A space is connected if there is no separation.
Otherwise, it is disconnected.
Example of separation
Let X be an arbitrary set with trivial topology. (The only open sets are \emptyset and X.)
This space is connected.
Let X=\{a,b\} with discrete topology. Then X is disconnected.
A separation is given by U=\{a\} and V=\{b\}.
Theorem of separation and clopen sets
Note that U,V give a separation of X if and only if U=V^c and V is open, then U is closed and open.
So if X is connected, then there is a non-empty proper (not the same as X) closed and open set.
The reverse is also true. (If the only clopen sets are \emptyset and X, then X is connected.)
Example of connected and disconnected space in real numbers
Let X=[a,b] with subspace topology inherited from \mathbb{R} is connected.
Then other connected subspace of \mathbb{R} are (a,b), [a,b), (a,b], (-\infty,b), (-\infty,b], (a,\infty), [a,\infty), and \mathbb{R}.
If X\subseteq \mathbb{R} with the subspace topology such that there are a<b<c with a,c\in X, b\notin X, then X is disconnected.
Note that X=((-\infy,b)\cap X)\cup ((b,\infty)\cap X) are two disjoint open sets whose union is X.
U is not empty because a\in U.
V is not empty because c\in V.
U\cap V=\phi because b\notin U\cap V, is a valid separation of X.
So X is disconnected.
Definition of totally disconnected space
Any topological space that any subset of it with at least two elements is disconnected is called a totally disconnected space.
Example of totally disconnected space
In \mathbb{R}, any subset of rational numbers with at least two elements is disconnected.
Because there is a irrational number between any two rational numbers.
Example of disconnected space
Let X\subseteq \mathbb{R}^2 and X=U\cap V, where U=\{(x,y)\in \mathbb{R}^2\mid y=1/x\} and V=\{(x,y)\in \mathbb{R}^2\mid x=0\}.
Then X is disconnected since U, V gives a separation of X (In this case, U and V are closed sets in \mathbb{R}^2).
Lemma of separated subsets
Let U,V give a separation of a topological space X. Let Y\subseteq X with the subspace topology is connected. Then Y is either contained in U or V.
Proof
Consider U'=U\cap Y and V'=V\cap Y. Then U' and V' are disjoint nonempty open subsets of Y. and Y=U'\cup V'.
Since Y is connected, then U' or V' are not a separation, so U' or V' is empty.
Theorem of connectedness of union of connected subsets
Let X=\bigcup_{\alpha\in I} X_\alpha such that \bigcap_{\alpha\in I} X_\alpha is non-empty. And X_\alpha are connected. Then X is also connected.
Proof
Let x\in \bigcap_{\alpha\in I} X_\alpha. By contradiction, suppose U,V give a separation of X. Assume x\in U and x\notin V, Applying the lemma to Y=X_\alpha for each \alpha\in I, we have X_\alpha\subseteq U or X_\alpha\subseteq V.
Since x\in X_\alpha is an element of u, the fist possibility holds, so \bigcap_{\alpha\in I} X_\alpha\subseteq U implies X\subseteq U, then U=x, V=\emptyset, which is a contradiction.
Example
Let X=S^1\subseteq \mathbb{R}^2 with the subspace topology. Let X_0=S^1\cap \{(x,y)\mid x\leq 0\} and X_1=S^1\cap \{(x,y)\mid x\geq 0\}.
Then X_0\cap X_1=\{(0,1), (0,-1)\}.
Note that both of them are homeomorphic to [0,1]\subseteq \mathbb{R}, which are known to be connected.
Proposition of connectedness and homeomorphism
Connectedness is a topological property (preserve under homeomorphism).
i.e. If X and Y are homeomorphic, then X is connected if and only if Y is connected.
Proof
By contradiction, U,V give a separation of X let \phi:X\to Y be a homeomorphism. Then \phi(U) and \phi(V) are disjoint nonempty open subsets of Y whose union is Y.
So Y is disconnected.
This contradicts the assumption that Y is connected.
Therefore, X is connected.
The reverse direction is similar.
Note
The connectedness of two topological spaces can be preserved under weaker conditions than homeomorphism.
Proposition of connectedness and continuous map
If X is connected and f:X\to Y is a continuous map, then f(X)\subseteq Y with subspace topology is connected.
Proof
By contradiction, suppose f(X) is disconnected. Then there are disjoint nonempty open subsets U,V of f(X) such that f(X)=U\cup V.
Since f is continuous, f^{-1}(U) and f^{-1}(V) are open in X and X=f^{-1}(U)\cup f^{-1}(V).
Since X is connected, then f^{-1}(U) and f^{-1}(V) are not a separation, so f^{-1}(U) or f^{-1}(V) is empty.
This contradicts the assumption that X is connected.
Therefore, f(X) is connected.