Math4201 Topology I (Lecture 9)

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 $(X,\mathcal{T})$.

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.

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, $x=y$.

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.

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.

$\Rightarrow$:

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 $X$.

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.

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 $X$)

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).