Topology: is this path connected?

[quasar987]

Homework Statement

Consider the sets A=\{(t,\sin(1/t))\in \mathbb{R}^2:t\in(0,1]\}, B=\{(0,s)\in\mathbb{R}^2:s\in[-1,1]\}. Let X=A\cup B. We consider on X the topology induced by the open ball topology of R².

a) Is X connected?

b) Is X path connected?

The Attempt at a Solution

a) I found that it is connected.

b) I concluded that it is not path-connected but it's a little touchy and I want a second opinion.

In case you haven't visualized the set yet, it consists of the union of the vertical line \{0\} \times [-1,1] with a sine wave on (0,1] that oscillates faster and faster as t-->0.

There is of course no difficulty in connecting two points of A or two points of B. The interesting case is when one tries to connect a point of A with one of B. Without loss of generality, let's try to connect (1,sin(1)) with (0,0). The only way is to follow the sine wave:

\gamma:[0,1]\rightarrow X[/itex]<br /> <br /> \gamma(t)=\left\{\begin{array}{cc}(0,0)&amp;amp;\mbox{if} \ \ &lt;br /&gt; t=0\\(t,\sin(1/t)) &amp;amp; \mbox{if} \ \ t\in(0,1]\end{array}<br /> <br /> Is gamma continuous? Let&#039;s consider an open ball of radius, say, ½, centered at (0,0). The intersection of that ball with X is an open in X that we will call O. Let&#039;s look at the pre-image of O by gamma. It contains {0} and an infinity of open intervals of (0,1] but \gamma^{-1}(O) is not open because it is impossible to find an open nbh of {0} that be entirely in \gamma^{-1}(O).<br /> <br /> N.B. We haven&#039;t covered caracterisation of continuity in metrizable spaces, so I can&#039;t use that.

 

[StatusX]

You almost got it. First of all, don't try to write an explicit formula for gamma. Your argument needs to apply to any possible path, not just that one. Next, what does this mean?

It contains {0} and an infinity of open segments of (0,1] but \gamma^{-1}(O) is not open because it is impossible to find an open nbh of (0,0) that be entirely in \gamma^{-1}(O).

I think you have the right idea, but its a little confusing. In the last part, are you talking about (0,0) or its preimage? I think you mean the latter, but you haven't rigorously argued why this should be true.

 

[quasar987]

Yes, I've edited the OP: (0,0) became {0} and "open segments" became "open intervals".

 

[quasar987]

What I've shown but not written in my post is that the path connecting (0,0) to (1,sin(1)) must take every value of (t,sin(1/t)) for t in (0,1] otherwise the path is not continuous. So the gamma I've written up there can be used without loss of generality because at most, another path would have points on the little vertical line \{0\} \times [-1,1], but this is irrelevant to my argument.

Also, it is easy but tiresome to show that given any radius r for an open-ball nbh of {0}, I can identity an interval [a,b] with 0<a<b<r that corresponds to two consecutive zeroes of sin(1/t), and I can find a sub-interval [c,d] with a<c<d<b for which the points (t,sin(1/t)) are outside of O.

 

[StatusX]

What I've shown but not written in my post is that the path connecting (0,0) to (1,sin(1)) must take every value of (t,sin(1/t)) for t in (0,1] otherwise the path is not continuous. So the gamma I've written up there can be used without loss of generality because at most, another path would have points on the little vertical line \{0\} \times [-1,1], but this is irrelevant to my argument.

Fine, but just because it passes through all those points doesn't mean it's the function you wrote, which passes through them without changing direction and at constant speed. We're really arguing over nothing though, because you never use that function in the rest of your proof, so all you need to do is erase it (and say what you said above about passing through all the points).

Also, it is easy but tiresome to show that given any radius r for an open-ball nbh of {0}, I can identity an interval [a,b] with 0<a<b<r that corresponds to two consecutive zeroes of sin(1/t), and I can find a sub-interval [c,d] with a<c<d<b for which the points (t,sin(1/t)) are outside of O.

What you said here is plenty to show it. I don't know what else you'd need to say without being needlessly detailed. It's just that what you had before was not enough.

 

[quasar987]

Cool, thanx.