Path-connected sets and path components

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Problem 1.6.6 from Lawson T. "Topology: A Geometric Approach" [Law06]More problems in Section 1.6 "Connectedness"

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.

Problem 1.6.7 from Lawson T. "Topology: A Geometric Approach" [Law06]More problems in Section 1.6 "Connectedness"

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

Remark.

Let $z$ be the common point. Then show that given $x, y$ in the union, we can find a path that joins them by using paths in individual path connected spaces that join $x$ to $z$ and join $z$ to $y$ .

Problem 1.9.46 from Lawson T. "Topology: A Geometric Approach" [Law06]More problems in Section 1.9 "Supplementary exercises"

Consider the space $X = A \cup B$ , where $A = \left\{ (x, \sin 1/x) : 0 < x \leq 1\right\}$ and $B = \left\{ (0, y) : -1 \leq y \leq 1\right\}$ (the topologist's sine curve; see the figure in Exercise 1.9.44). We show here that $A \cup B$ is not path connected. Suppose $A \cup B$ were path connected. Let $f : [0, 1] \to A \cup B$ be a path with $f(0) = (0, 0)$ and $f(1) = (1, \sin 1)$ . Consider the set $S = \left\{ t : f([0, t]) \subset B\right\}$ . Let $p : \mathbb {R}^{2} \to \mathbb {R}$ be $p(x, y) = x$ .

(a)

Show that $S$ is nonempty, is bounded from above, and has a least upper bound $u < 1$ .

(b)

Show that $u \in S$ .

(c)

Show that there is a neighborhood $N_{f(u)}$ of $f(u)$ consisting of an infinite number of separated arcs, and a neighborhood $N_u$ of $u$ with $f(N_u) \subset N_{f(u)}$ .

(d)

Show that there is $u_1 > u$ , $u_1 \in N_u$ with $pf(u_1) > 0$ .

(e)

Show that there are disjoint open sets $U, V \subset N_{f(u)}$ with $f(u) \in U$ , $f(u_1) \in V$ , and $U \cup V = N_{f(u)}$ ; that is, $N_{f(u)}$ is separated by $U, V$ .

(f)

By looking at $f|_{[u, u_1]}$ , arrive at a contradiction.

References

[Law06] Lawson T. Topology: A Geometric Approach. Oxford University Press, 2006.