Topology: An Inquiry-Based Approach

Find a topological space $X$ and a point $x\in X$ such that the minimal neighborhood of $x$ is not an open set.

In Activity 18.7 of Chapter 18 we showed that an arbitrary union of connected sets is connected provided the intersection of those sets is not empty. Is the same result true for path connected sets. That is, if $X$ is a topological space and $\{A_{\alpha }\}$ for $\alpha$ in some indexing set $I$ is a collection of path connected subsets of $X$ and $\bigcap _{\alpha \in I}{A}_{\alpha }\ne \mathrm{\varnothing }\text{,}$ must it be the case that $A=\bigcup _{\alpha \in I}{A}_{\alpha }$ is path connected? Prove your answer.

In Exercise 4 we see an example of a space that is path connected but not locally path connected. Is it possible to find a space that is locally path connected but not path connected? Verify your answer.

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 not path connected. (Hint: Suppose there is a path between $a$ and $b$ where $a<0$ and $b>1\text{.}$)

We know that a space can be connected but not path connected. We also know that local path connectedness does not imply connectedness. However, if we combine these conditions then a space must be path connected. That is, show that if a topological space $X$ is connected and locally path connected, then $X$ is path connected.