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Math4201 Topology I (Lecture 23)
Connectedness of topological spaces
Connected space
Definition of connected space
Let X be a topological space. X is separated if there exist two disjoint nonempty open subsets U,V\subset X such that X=U\cup V.
If X is not separated, then X is connected.
Any interval in \mathbb{R} with standard topology is connected
Let I=[a,b] be an interval in \mathbb{R} with standard topology. Then I is connected.
Proof
By contradiction, we assume that U,V give a separation of \mathbb{R}. In particular, \exists a\in U, b\in V.
Let a_0\coloneqq \sup\{x\in U\cap [a,b]\}. Note that a\in U\cap [a,b], so a_0\geq a. Since any element of U\cap [a,b] is less than or equal to b, a_0\leq b.
Case 1: a_0=a
Since U is open, there is \epsilon>0 such that [a,a+\epsilon)\subset U\cap [a,b]. So a_0\geq a+\epsilon>a, which contradicts the definition of a_0.
Case 2: a_0=b.
Since V is open, there is \epsilon>0 such that (b-\epsilon,b]\subseteq V\cap [a,b]. This implies that b-\epsilon is also an upper bound of U\cap [a,b], so a_0\leq b-\epsilon<b, which contradicts the definition of a_0.
Case 3: a<a_0<b.
Subcase I: a_0\in U.
There is an \epsilon>0 such that (a_0-\epsilon,a_0+\epsilon)\subset U\cap [a,b] because U is open.
In particular, a_0 is greater than any element of (a_0-\epsilon,a_0+\epsilon), which contradicts the definition of a_0.
Subcase II: a_0\in V.
There is an \epsilon>0 such that (a_0-\epsilon,a_0+\epsilon)\subset V\cap [a,b] because V is open.
In particular, a_0 is less than any element of (a_0-\epsilon,a_0+\epsilon). Since a_0 is an upper bound of U\cap [a,b], any point >a_0 is not in U\cap [a,b].
So, U\cap [a,b]\subseteq [a,a_0-\epsilon). This shows that a_0-\epsilon is an upper bound of U\cap [a,b], which contradicts the definition of a_0.
Intuitively, since both sets in \mathbb{R} are open, you cannot set a clear boundary between the two sets by least upper bound argument.
Corollary as Intermediate Value Theorem
If f:[a,b]\to \mathbb{R} is continuous, and c\in\mathbb{R} is such that f(a)<c<f(b), then there exists x\in [a,b] such that f(x)=c.
Proof
Since [a,b] is connected, since f is continuous, f([a,b]) is connected.
By contradiction, if c\notin f([a,b]), then f([a,b]) has two points f(a),f(b) and c is a point between that isn't in f([a,b]). This contradicts the connectedness of f([a,b]).
So f(a)<c<f(b) or f(b)<c<f(a) must hold.
Definition of path-connected space
A topological space X is path-connected if for any two points x,x'\in X, there is a continuous map \gamma:[0,1]\to X such that \gamma(0)=x and \gamma(1)=x'. Any such continuous map is called a path from x to x'.
Note
Path-connectedness is a stronger condition than connectedness.
Theorem of path-connectedness and connectedness
Any path-connected space is connected.
Proof
By contradiction, let U,V be a separation of X. In particular, \exists x\in U, x'\in V.
Since X is path-connected, \exists \gamma:[0,1]\to X such that \gamma(0)=x and \gamma(1)=x'.
Then since \gamma is continuous, \gamma^{-1}(U) and \gamma^{-1}(V) are open in [0,1] and [0,1]=\gamma^{-1}(U)\cup \gamma^{-1}(V). We want to show that this gives a separation of [0,1].
Since U\cap V=\emptyset, \gamma^{-1}(U) and \gamma^{-1}(V) are disjoint.
U\cup V=X so \gamma^{-1}(U)\cup \gamma^{-1}(V)=[0,1].
Each of \gamma^{-1}(U) and \gamma^{-1}(V) is non-empty because x\in U\implies 0\in \gamma^{-1}(U) and x'\in V\implies 1\in \gamma^{-1}(V).
This contradicts the assumption that [0,1] is connected.
Example of path-connected space
A subspace X of \mathbb{R}^n is convex if for any two points x,x'\in X, the line segment connecting x and x' is contained in X.
In particular B_R(x) is convex. So X is path-connected.
Let X=\mathbb{R}^n\setminus\{0\}. with n\geq 2. Then X is path-connected. (simply walk around the origin)
Theorem for invariant property of connectedness
If f:X\to Y is a continuous and surjective map, and X is connected, then Y is connected.
Proof
Take y,y'\in Y, since f is surjective, \exists x,x'\in X such that f(x)=y and f(x')=y'. Let \gamma:[0,1]\to X be a path from x to x'.
Then f\circ \gamma:[0,1]\to Y is a continuous map. and f\circ \gamma(0)=y and f\circ \gamma(1)=y'.
Example of connected but not path-connected space
Let A=\{(x,y)\in \mathbb{R}^2\mid y=\sin(1/x), x>0\}. Then A is connected, and also path-connected.
However, take \overline{A}=A\cup \{0\}\times [-1,1]. Then \overline{A} is not path-connected but connected.
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