Topology: An Inquiry-Based Approach

Recall from Definition 12.13 that if ${\tau }_1$ and ${\tau }_2$ are two topologies on a set $X$ such that ${\tau }_1\subseteq {\tau }_2\text{,}$ then ${\tau }_1$ is said to be a coarser (or weaker) topology than ${\tau }_2\text{,}$ or ${\tau }_2$ is a finer (or stronger) topology than ${\tau }_1\text{.}$ In this exercise we explore the question of whether compactness is a property that is passed from weaker to stronger topologies or from stronger to weaker. Let ${\tau }_1$ and ${\tau }_2$ be two topologies on a set $X\text{.}$ If ${\tau }_1\subseteq {\tau }_2\text{,}$ what does connectedness under ${\tau }_1$ or ${\tau }_2$ imply, if anything, about compactness under the other topology? Justify your answers.

Let $f:[a,b]\to \mathbb{R}$ be a continuous function from a closed interval into the reals. Let $U=f(u)$ and $V=f(v)$ be such that $U\le f(x)\le V$ for all $x\in [a,b]\text{.}$ Prove that there is a $w$ between $u$ and $v$ such that $f(w)(b-a)={\int }_a^{b}f(t)dt\text{.}$

Let $X=\mathbb{R}$ with the finite complement topology. We have shown that every subset of any topological space with the finite complement topology is compact. Now find all of the connected subsets of $X\text{.}$ Prove your result.

Let $f:S^1\to \mathbb{R}$ be a continuous function. Show that there is a point $x\in S^1$ with $f(x)=f(-x)\text{.}$

Let $a,b\in \mathbb{R}$ with $a<b\text{.}$ Explain why no two of the sets $(a,b)\text{,}$ $(a,b]\text{,}$ and $[a,b]$ homeomorphic.

Let $K=\{\frac{1}{k}\mid k\text{ is a positive integer}\}\text{.}$ Let $\mathcal{B}$ be the collection of all open intervals of the form $(a,b)$ and all sets of the form $(a,b)\setminus K\text{,}$ where $a<b$ are real numbers as in Example 13.13. Let ${\tau }_K$ be the topology generated by $\mathcal{B}\text{.}$ Show that $(\mathbb{R},{\tau }_K)$ is a connected space.

Let $a,b\in \mathbb{R}$ with $a<b\text{.}$ Prove that a homeomorphism $f:[a,b]\to [a,b]$ carries end points into end points.

Let $X$ be a topological space with only a finite number of connected components. Show that each component of $X$ is open.

Let $X$ and $Y$ be connected spaces with $f:X\to Y$ a continuous function. Is it the case that if $S$ is a cut set of $X\text{,}$ then $f(S)$ is a cut set of $Y\text{?}$ Prove your answer.