Math4201 Topology I (Lecture 23)

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 $a_0$.

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.

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.

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)

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 $f\circ \gamma(1)=y'$.

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.

Show next time