Path-connected/connected space

[Incand]

Homework Statement

Show that
$S = \{(x,y):0 < x \le 1, y=\sin (1/x)\} \cup \{(x,y): -1 \le x \le 0, y= 0\}$
is connected but not path connected in $\mathbf R^2$.

Homework Equations

[/B]
Let $X$ be a metric space.
A set $E\subseteq X$ is said to be connected if $E$ is not a union of two nonempty separated sets.

Two subsets $A,B$ of $X$ are said to be separated if $\bar A \cap B = A \cap \bar B = \emptyset$

A set $S$ in $\mathbf R^n$ is said to be path connected if there for every pair of points $a,b \in S$ there is a continuous function $f:[0,1] \to S$ so that $f(0) = a, f(1)=b$.

3. The Attempt at a Solution
The way forward seems to be to assume $S$ is not connected, i.e. $S=A\cup B$ where $A$ and $B$ are seperated. Now I need to find a contradiction but I don't even know where to start, any hints?

As for proving it's not path connected we need to select two promising points $a,b$. Clearly we need to select $a \in [-1,0]$ and $b\in (0,1)$ (or the opposite) or they're obviously path connected. Then I think it's easiest to assume that it is path connected and that there is a function $f$ satisfying the above and then get a contradiction again.

Again I'm quite clueless how to get started. As I remember it a vector valued function is continuous iff every coordinate is continuous. Initially I thought I only had to look at the $y$ coordinates in which case one can actually find such a continuous function by choosing $\xi \in (0,b]$ that satisfy $f(\xi) = 0$
And then define
$g(t) = \begin{cases} a+t \text{ for } t \in [0,-a]\\ b -\frac{(b-\xi)(1-t)}{1+a} \text{ for } t > -a \end{cases}$
and use that as our $x$ coordinate. Luckily in this case $x([0,1]$ isn't continuous so at least I didn't prove the impossible.

 

[pasmith]

Both subsets in the definition are graphs of continuous functions, so each on its own is connected (and path connected).

Thus for connectedness of the union you must ask the following question:

Let $\epsilon > 0$. Does the open ball of radius $\epsilon$ whose centre is at $(-\frac12,0)$ necessarily contain points in $\{(x, \sin x^{-1}) : 0 < x \leq 1\}$?

 

[Incand]

Why $(-1/2,0)$? You don't mean $(0,0)$?. The first open ball obviously doesn't contain any points in the set for $\epsilon \le 1/2$ at least.

The open ball around $(0,0)$ should contain points in the set. We need points satisfying $\sqrt{x^2+\sin^2 1/x } < \epsilon$ using the euclidian metric. However for each $\epsilon > 0$ we can choose $x = (\pi n)^{-1}< \epsilon$ for some $n \ge N$ since the sequence converges. This $x$ then is in the set and is also in the ball of radius $\epsilon$.

 

[fresh_42]

Try to understand what it is about first, as @pasmith suggested you to do. The goal of this example is to understand the concepts. Technical details are the minor part of it. So think about the idea behind it, and put your reasoning into a proof afterwards.

Both sets are separated by a gap at $(0,-)$, so a path from left to right would lead to a discontinuity. Why? Can't we bypass the gap $\bumpeq$? Esp. what did you mean by $a\in [0,1]$? Your sets are points in the plane, so what is $a$? "two promising points" isn't a good guide towards rigor. Simply assume a path from left to right and show why "one promising" point $(0,\text{ with anything })$ leads to a contradiction. How does a path look like in terms of functions?

Now why can they despite of this be still connected? What's the difference in the concepts?

 

[Incand]

I'm sorry but there was a typo in my initial post. It should be $\{(x,y): -1 \le x \le 0, y=0 \}$ i.e. also equal to zero (otherwise it wouldn't have been connected). I fixed that in the initial post now.

Lets name the set with the $\sin$ $A$ and the other one $B$.

Well the problem with the gap is essentially that $\sin 1/x$ doesn't converge. We have that $(0,[-1,1])$ are limit points of $A$ in addition to the set itself. Since $(0,0)$ is a limit point (of both sets) it means that no function passing through this point is continuous since $\sin 1/x$ doesn't have a limit.

As for connected $A$ does have a limit point in $B$ so they are arbitrary close.

Does this capture the concept?

 

[fresh_42]

I'm sorry but there was a typo in my initial post. It should be $\{(x,y): -1 \le x \le 0, y=0 \}$ i.e. also equal to zero (otherwise it wouldn't have been connected). I fixed that in the initial post now.

Lets name the set with the $\sin$ $A$ and the other one $B$.

Well the problem with the gap is essentially that $\sin 1/x$ doesn't converge. We have that $(0,[-1,1])$ are limit points of $A$ in addition to the set itself. Since $(0,0)$ is a limit point (of both sets) it means that no function passing through this point is continuous since $\sin 1/x$ doesn't have a limit.

As for connected $A$ does have a limit point in $B$ so they are arbitrary close.

Does this capture the concept?

Yes it does. Just one thing: If we have $(0,0)$ as limit point of both, why can't we find a path through this point as a gate between the two?
For the path connection argument, you could formally assume a path $p: [0,1] \rightarrow S=A\cup B$ with $p(0)\in B\, , \,p(1)\in A\; , \;p=(p_1,p_2).$ Then for $p$ to be continuous, both components $p_i$ have to be. Now the question is, why can't there be a path $p_2$ between $p_2(1)=\sin(\frac{1}{x(1)})$ and $p_2(t_0)=0$ (although it's a limit point), and now the continuity argument applies to $p_2(t_0)=\sin\frac{1}{t_0}$ alone. But this only formalizes the step from a path in the plane to the properties of the single-valued sine function.

 

[Incand]

Yes it does. Just one thing: If we have $(0,0)$ as limit point of both, why can't we find a path through this point as a gate between the two?.

We have this useful theorem stating that if $p$ is a limit point of $E\subseteq X$ then $f$ is continuoius if and only if $\lim_{x\to p} =f(p)$.
(For $f:E \to Y$ for metric spaces $X,Y$, $p\in E$).

And since the limit doesn't exists the function can't be continuous in $(0,0)$ either. Where we considered the $y(x)$ coordinate and since $y(x)$ isn't continuous neither is the vectorvalued function $f(x) = (x,y(x))$.

And we have that $\sin 1/x(t)$ isn't continuous at $x(t) = 0$,
As for that there isn't any other ''jump point'' between $A$ and $B$ it's easy to see that for any point $p\in A$ $\exists \epsilon > 0$
$\inf_{z\in A} d(z,p) > \epsilon$ and similarly for $p\in B \backslash (0,0)$, $\inf_{z \in B} d(z,p) > \epsilon$.

Putting together all we done so far I think that proves that $S$ isn't path-connected?

 

[Incand]

As for them being connected I think there's still work to do. @pasmith suggested that I show that $(0,0)$ is a limit point of $A$ which we did. (If he/she meant (0,0)?)

As to turn that into a proof that $S$ is connected seems harder. Obviously we have that $A,B$ aren't separated.
Two separated sets must also be disjoint so they can't share a point. And every point in $A$ and $B$ is a limit point so having a 'gap' between the sets there doesn't work. How do I turn this into a real proof?

 

[fresh_42]

As for them being connected I think there's still work to do. @pasmith suggested that I show that $(0,0)$ is a limit point of $A$ which we did. (If he/she meant (0,0)?)

As to turn that into a proof that $S$ is connected seems harder. Obviously we have that $A,B$ aren't separated.
Two separated sets must also be disjoint so they can't share a point. And every point in $A$ and $B$ is a limit point so having a 'gap' between the sets there doesn't work. How do I turn this into a real proof?

Well, since path connection implies connection, $A$ and $B$ are the only candidates, or at least their common boundary $\{(0,0)\}$. So $\overline{A}\cap B \neq \emptyset$. Do I miss something here?

 

[Incand]

Well, since path connection implies connection, $A$ and $B$ are the only candidates, or at least their common boundary $\{(0,0)\}$. So $\overline{A}\cap B \neq \emptyset$. Do I miss something here?

I'm sure you don't but I'm not sure how everything follows from this. Why are $A$ and $B$ the only candidates? I'm with you that $A$ and $B$ are each a connected set but how does this show that every other choice of separated sets doesn't work?

If I take two sets so that their union is $S$ I could probably show they're not separated but there is an infinite number of those sets and I need to show it for all of those.

 

[fresh_42]

If $S=A'\cup B'$ with separated sets $A',B'$ and $(0,0)$ is a inner point of one of them, then $S$ would be path connected, because $(0,0)$ could be chosen as an inner point of some path, and we could extend this path to one, that connects all points in $S$.
So $(0,0)$ has to be a boundary point (and within one of the sets), say of $B'$. But then every open neighborhood of $(0,0)$ contains also elements of $A'=S-B'$ and is then a limit point of $A'$, i.e. $(0,0)\in \overline{A'} \cap B'.$

(But please check this argument, I tend to rely on intuition which isn't a good adviser, if it comes to topology.)

 

[pasmith]

Why $(-1/2,0)$? You don't mean $(0,0)$?.

It should be (0,0), yes; I misread the set in question as the standard "graph of sin 1/x together with a vertical line on the x-axis", but the question actually specifies a horizontal line.

 

[Incand]

If $S=A'\cup B'$ with separated sets $A',B'$ and $(0,0)$ is a inner point of one of them, then $S$ would be path connected, because $(0,0)$ could be chosen as an inner point of some path, and we could extend this path to one, that connects all points in $S$.
So $(0,0)$ has to be a boundary point (and within one of the sets), say of $B'$. But then every open neighborhood of $(0,0)$ contains also elements of $A'=S-B'$ and is then a limit point of $A'$, i.e. $(0,0)\in \overline{A'} \cap B'.$

(But please check this argument, I tend to rely on intuition which isn't a good adviser, if it comes to topology.)

That argument seems fine as far as I can tell. Nice work! So this shows (0,0) can't be part of either set.

Another thing to note is that if $S=A'\cup B'$ we must have $A \subseteq A'$ or $B \subseteq B'$ (or the other way around) since if $A'$ contains a proper subset $E\subset A$ and $(A-E) \subset B$ they clearly share a limit point since every point of $A$ and $B$ respectively is a limit point. Sets like these was what initially worried me even though it's obvious that they're not separated.