Math4201 Topology I (Lecture 22)

Let $X$ be an arbitrary set with trivial topology. (The only open sets are $\emptyset$ and $X$.)

This space is connected.

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Let $X=\{a,b\}$ with discrete topology. Then $X$ is disconnected.

A separation is given by $U=\{a\}$ and $V=\{b\}$.

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 $\mathbb{R}$.

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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.

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.

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 $\mathbb{R}^2$).

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.

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.

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.

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.

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.