# connected but not path connected?

Consider the following subsets of $\mathbb R^2$.

A=$\{(x,y); 0\le x \le 1, y=\frac xn$ for $n \in \mathbb N\}$

B=$\{(x,y); \frac12 \le x \le 1, y=0\}.$

How do I prove that $A\cup B$ is connected but not path connected.

I really have no idea how to solve this problem. I stared at it for many hours, tried different ridiculous things, but just can't figure out. Thank you very much in advance.

• You probably mean $0\leq x\leq1$ and $\frac{1}{2}\leq x \leq 1$? – Listing Mar 18 '14 at 9:40
• Can't you just apply the definitions? – Sylvain Biehler Mar 18 '14 at 9:41
• $\frac{1}{2}\ge 1$ !!!??? – Semsem Mar 18 '14 at 9:47

You have to prove $2$ things here.
Not path connected: To prove that $A\cup B$ is not path connected, you must find $2$ points for which a path between them does not exist. Since it is easy to see that $A$ and $B$ are path connected, it is best to take one point from $A$ and one from $B$. For example, take $$a=(0,0)\in A,\\b=(1, 0)\in B.$$ Now take any path $$\gamma:[0,1]\to A\cup B$$ with $\gamma(0)=a, \gamma(1)=b$ and try to show that $\gamma$ cannot be continuous. For this, take a look at the last point $t\in[0,1]$ for which $\gamma(t)$ is still on $A$, and show that there exists a $\delta>0$ such that for any $\epsilon>0$, the point $\gamma(t+\epsilon)$ is not $\delta$-close to $\gamma(t)$.
Connected: To prove that $A\cup B$ is connected, you must show that open and disjoint nontrivial sets $U,V$ for which $U\cup V=A\cup B$ do not exist, or, equivalently, that there does not exist a nontrivial set $W\subset A\cup B$ which is both open and closed. I suggest you use this second definition and takt a set $W$ which is closed and opened in $A\cup B$.
Without loss of generality, you can assume that it contains at least one point in $A$ (otherwise, take the complement of $W$). Now, because $A$ is connected, you can show that $A\subseteq W$. Using the fact that $W$ is closed, meaning that $W$ contains the closure of $A$ in $A\cup B$, should tell you that $W$ contains $B$ as well, meaning $W=A\cup B$.