# the topologist's sine circle is path-connected but it's not locally path-connected

As you may know the topologist's sine curve is the set:

$$\{{(x,y) : x=0 \ and \ |y|\leq 1,\ or \ 0

I want to show that the topologist's sine circle which is the union of circular arc and topologist's sine curve is path-connected but it's not locally path-connected. (you can see the picture of topologist's sine circle below):

[topologist's sine circle]

for saying that it is not locally path-connected I think we have to choose an interval near the $$x=0$$ and then show that it's not locally path-connected. Also the path-connectedness of this object is clear from the image, but unfortunately I don't know how to make a mathematical proof for being path-connected and not being locally path-connected.

• BTW this space is also called Warsaw circle. Google Images, Google, StackExchange. – Martin Sleziak May 8 '15 at 9:19
• @MartinSleziak I searched it and I found the solution. thanks for your help :) – F.K May 8 '15 at 9:41
• F.K.: If you found somewhere online a link which answers your question, you could at least post it in a comment or, even better, post your own answer to the question and include the comment there. It is good to make this post useful also for other people which come here to ask the same question. – Martin Sleziak May 8 '15 at 9:49
• BTW I think that in general it is recommended to add pictures using built-in editor, which creates a permanent imgur link - like this i.stack.imgur.com/dLjny.jpg - rather then outside source - like s10.postimg.org/osrpohjc9/topologist_s_sine_circle.jpg in your post. See, for example, here: meta.math.stackexchange.com/questions/4205/… – Martin Sleziak May 9 '15 at 5:20

This space is path connected.

• You have a continuous path joining $(0,0)$ and $(1,0)$, namely the arc added to the topologist's sine curve.
• There is a continuous path joining $(0,0)$ with any point of the segment $\{0\}\times[-1,1]$, you just take a straight line.
• There is a continuous path joining $(1,0)$ with any point $(x,\sin\frac1x)$, $x>0$. You can simply take part of the graph of continuous function $x\mapsto \sin\frac1x$

Now you can simply combine the above to get a path joining any two points. Or you can argue that the above implies that any two points lie in the same path-connected component.

To show that this space is not locally path-connected, you can simply take small enough neighborhood of the point $(0,1)$. Intersection of such neighborhood is not even connected. (This is basically the same as the argument why topologist's sine curve is not locally connected.)

It is worth mentioning that this space is also called Warsaw circle. Google Images, Google, StackExchange.

• I think if $x \neq 0$ there is a path between $(1,0)$ and $(x,sin\dfrac{1}{x})$. there is no need for $x$ to be positive. – F.K May 8 '15 at 11:11
• Well, since in your space you only have points of the form $(x,\sin 1/x)$ for $x\ge0$, writing $x>0$ and $x\ne0$ is, in this case, the same thing. – Martin Sleziak May 8 '15 at 11:17