# $\sin(1/x)$ space with a line is not locally path connected?

Let $X = \text{$\sin(1/x)$space } \cup \text { line from$(0,1)$to$(\frac{1}{2 \pi}, 0$})$,

Where $$\sin(1/x) \text { space }= \{(0,y) : y \in [-1,1] \cup \{(x, \sin(1/x)) : x \in (0, \frac{1}{2 \pi}]\}$$

Show that $X$ is path connected but not locally path connected.

The definition of locally path connected: $X$ is locally path connected if, for each $x \in X$ and every open neighborhood $U$ of $x$, there is an open $V$ with $x \in V \subseteq U$ such that any two points in $V$ can be jointed by a path in $U$.

Path connected is simple to prove, but I can't see how this isn't locally path connected.

Anyone have any ideas?

• Where did you get stuck when you looked at the definition of locally path connected and attempted to prove it? – Lee Mosher May 16 '17 at 13:27
• The point $(0,-1)$ has no path to any point on the graph. – Clayton May 16 '17 at 13:28
• @Clayton But if the space is path connected, how can $(0,-1)$ have no path to other points on the graph? – Oliver G May 16 '17 at 13:31
• @Clayton It has a path, just not a local one... – SEWillB May 16 '17 at 13:31
• @TonyK: Of course not. Why would you think that? Path-connectedness and local path-connectedness are mostly orthogonal properties (and the Warsaw circle is probably the simplest example of that). – tomasz May 16 '17 at 14:01

It is probably easiest to show that the point $(0, 0)$ doesn't satisfy the property of being locally path connected. The picture below shows the space in question: (ignore the left of the y-axis) Pick an open ball of radius $R$ about $(0,0)$ with $R$ small enough that the ball doesn't intersect the green line. Denote this $B((0,0), R)$. Given any open subset $V$ of $B((0,0), R)$ we must have an open ball of radius $r \le R$ with $B((0,0), r) \subset V$.
If we can show that these spaces are not path connected then we're done. However it is obviously the case that these spaces are not path connected. If we could join up the $\sin(1/x) \text{-space}$ to the $y$-axis by a continuous path all inside $B((0,0), r)$; then we could certainly join up simply the $\sin(1/x) \text{-space}$ to the $y$-axis not necessarily constrained to the open ball $B((0,0), r)$.