NoteNextra-origin/content/Math4201/Math4201_L9.md

Math4201 Topology I (Lecture 9)

Convergence of sequences

Let (X,\mathcal{T}) be a topological space and \{x_n\}_{n\in\mathbb{N}_+} be a sequence of points in X. We say x_n\to x as n\to \infty (x_n converges to x as n\to \infty)

if for any open neighborhood U of x, there exists N\in\mathbb{N}_+ such that \forall n\geq N, x_n\in U.

Example of convergence of sequences

Let X=\{a,b,c\} with the topology \mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}

Let x_n=b for all n\in\mathbb{N}_+. Then x_n\to b as n\to \infty.

Moreover, x_n\to a as n\to \infty since any open neighborhood of a (\{a,b\},${a,b,c}$) contains b.

Without loss of generality, x_n\to c as n\to \infty since any open neighborhood of c (\{b,c\},${a,b,c}$) contains b.

You may find convergence of sequences in more than one point.

Let x_n=a for all n\in\mathbb{N}_+. Then x_n\to a as n\to \infty (take U=\{a,b\})

A non-example of convergence of sequences:

Let x_n=\begin{cases}a, & n\text{ is even} \\ c, & n\text{ is odd}\end{cases}. Then x_n does not converge to any point in X. So this sequence does not have a limit in (X,\mathcal{T}).

More special topologies

Hausdorff space

A topological space (X,\mathcal{T}) is a Hausdorff space if for any two distinct points x,y\in X, there exist open neighborhoods U and V of x and y respectively such that U\cap V=\emptyset.

Non-example of Hausdorff space

Let X=\{a,b,c\} with the topology \mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}

Let x=a,y=b. Then they don't have disjoint open neighborhoods.

This topology is not a Hausdorff space.

If a topological space is a Hausdorff space, then every sequence has a unique limit point or have no limit point.

Properties of Hausdorff space

Let (X,\mathcal{T}) be a Hausdorff space. Then every sequence \{x_n\}_{n\in\mathbb{N}_+} converges to x and y, then x=y.

Tip

We want to show if x\neq y, then there exists an open neighborhood U of x and V of y such that U\cap V=\emptyset.

Proof

We proceed by contradiction.

Suppose x\neq y, then by definition of Hausdorff space, there exists an open neighborhood U of x and V of y such that U\cap V=\emptyset.

If x_n converges to x, then for any open neighborhood U_x of x, there exists N_x\in\mathbb{N}_+ such that \forall n\geq N_x, x_n\in U_x. Similarly, for any open neighborhood U_y of y, there exists N_y\in\mathbb{N}_+ such that \forall n\geq N_y, x_n\in U_y.

Then we can find N=max\{N_x,N_y\} such that x_n\in U_x\cap U_y for all n\geq N. This contradicts the assumption that U\cap V=\emptyset.

Therefore, x=y.

Properties of closed singleton

Let (X,\mathcal{T}) be a Hausdorff topological space. Then \forall x\in X, \{x\} is a closed set.

Non-example of closed singleton

Let X=\{a,b,c\} with the topology \mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}

Then \{b\} is not a closed set, since X\setminus \{b\}=\{a,c\} is not an open set.

Proof

We need to show that A=X\setminus \{x\} is an open set.

Take y\in A, then by the assumption, x and y have disjoint open neighborhoods U_x and V_y respectively. x\in U_x and y\in V_y and U_x\cap V_y=\emptyset.

So x\notin V_y, y\in V_y. So A\subseteq\bigcup_{y\in A,y\neq x} V_y.

Since \forall V_y,x\notin V_y, So A\subseteq\bigcup_{y\in A,y\neq x} V_y.

So A=\bigcup_{y\in A,y\neq x} V_y is an arbitrary union of open sets, so A is an open set.

Therefore, \{x\} is a closed set.

Continuous

Continuous functions

Definition for continuous functions

Let (X,\mathcal{T}) and (Y,\mathcal{T}') be topological spaces and f:X\to Y. We say that f is continuous if for every open set V\in Y, f^{-1}(V)\coloneqq\{x\in X: f(x)\in V\} is open in X.

That is, \forall V\in \mathcal{T}', f^{-1}(V)\in \mathcal{T}.

Definition for point-wise continuity

Let (X,\mathcal{T}) and (Y,\mathcal{T}') be topological spaces and f:X\to Y. We say that f is continuous at x\in X if for every open set V\in \mathcal{T}' such that f(x)\in V, there exists an open set U\in \mathcal{T} such that x\in U and f(U)\subseteq V. (f^{-1}(V) contains an open neighborhood of x)

Lemma for continuous functions

f:X\to Y is continuous if and only if \forall x\in X, f is continuous at x.

Proof

\Rightarrow:

Trivial

\Leftarrow:

Take an open set V\in \mathcal{T}'. Then for any point x\in f^{-1}(V), we have f(x)\in V.

In particular, by definition of continuity at x, there exists an open set U_x of x such that U_x\subseteq f^{-1}(V).

Then f^{-1}(V)=\bigcup_{x\in f^{-1}(V)} U_x is an arbitrary union of open sets, so f^{-1}(V) is open in X.

Example of continuous functions

Let X be any set and \mathcal{T} be the discrete topology on X, \mathcal{T}' be the trivial topology on X.

Let f:(X,\mathcal{T})\to (X,\mathcal{T}') be the identity function. Then f is continuous.

Since forall V\in \mathcal{T}', V is open in X, we can find f^{-1}(V) is open in X. (only neet to test X,\emptyset)

In general, if T is a finer than T', then f:(X,\mathcal{T})\to (X,\mathcal{T}') be the identity map is continuous.

However, if we let f:(X,\mathcal{T}')\to (X,\mathcal{T}) be the identity function, then f is not continuous unless X is a singleton.

Definition for constant maps

Let X,Y be topological spaces and y_0\in Y, f:X\to Y is defined by f(x)=y_0 for all x\in X. Then f is continuous.

Proof

Take an open set V\in \mathcal{T}'. f^{-1}(V)=\begin{cases}X, & y_0\in V \\ \emptyset, & y_0\notin V\end{cases} is open in X. (by definition of topology, X,\emptyset are open in X)

Example of inclusion maps

Let X be a topological space and A\subseteq X equipped with the subspace topology.

Let f:A\to X be the inclusion map f(x)=x for all x\in A.

Then take V\subseteq X be an open set. f^{-1}(V)=V\cap A\subseteq A is open in A (by subspace topology).