Connectedness

Show that a space $X$ is connected iff the only subsets of $X$ which are both open and closed are $\emptyset$ and $X$.

Proposition 1.6.1:

Consider the property: $(*)$ if $x, y \in A \subset \mathbb {R}$, then the interval $[x, y] \subset A$.

Proposition 1.6.3: A set $A \subset \mathbb {R}$ satisfies $(*)$ iff it is an interval, a ray, or $\mathbb {R}$.

It is straightforward to see that an interval, ray, or $\mathbb {R}$ satisfies $(*)$. For the converse, suppose $A$ satisfies $(*)$. The case where $A$ is bounded both from above and below has already been handled: let $a$ be the greatest lower bound and $b$ the least upper bound of $A$. This implies $A \subset [a, b]$. To show that $(a, b) \subset A$: let $c$ be a point in $(a, b)$. Since $a$ is the greatest lower bound, there is an element $e \in A$ with $a \leq e < c$. Similarly, $b$ being the least upper bound implies that there is an element $f \in A$ with $c < f \leq b$. But $(*)$ implies that $[e, f] \subset A$ and so $c \in A$. Hence $(a, b) \subset A$. But $A \subset [a, b]$, so there are four possibilities for $A$: $(a, b)$, $[a, b)$, $(a, b]$, $[a, b]$, each of which is an interval.

Complete the proof of the proposition by considering the other cases: where $A$ is not bounded on one side or the other or both.

Show that $\mathbb {R}$ is connected.

Proposition 1.6.6: Suppose that $A_i$ is a collection of connected subsets of a topological space $X$ so that they all have at least one point $a$ in common. Then the union $A = \cup_i A_i$ is connected.

Use this proposition to deduce that $\mathbb {R}$ is connected from the fact that a closed interval is connected.

A space $X$ is called path connected if, given $x, y \in X$, there is a continuous map $f : [0, 1] \to X$ (called a path in $X$) with $f(0) = x$ and $f(1) = y$. We say that the path connects $x$ to $y$.

Define a relation on $X$ by $x \sim y$ if there is a path connecting $x$ to $y$.

Show that the relation $x \sim y$ as defined above is an equivalence relation.

Show that a union of path connected sets with a point in common is path connected.

The intermediate value theorem states: suppose that $f : X \to \mathbb {R}$ is a continuous function and $X$ is connected. Let $c = f(x_1)$ and $d = f(x_2)$ and suppose that $c < e < d$. Then there is $x_3 \in X$ with $f(x_3) = e$.

Prove the intermediate value theorem using the fact that $f(X)$ is connected and the characterization of connected sets in the line (a subset of $\mathbb {R}$ is connected iff it is an interval, a ray, or $\mathbb {R}$).

Show that the letter T is not homeomorphic to the letter O.

Show that $S^1$ is not homeomorphic to $\mathbb {R}$ by showing $S^1 \setminus \left\{ x\right\}$ is not homeomorphic to $\mathbb {R} \setminus \left\{ y\right\}$.

Show that two disjoint concentric circles in the plane are not homeomorphic to one circle.