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Are the following sets connected or path connected?

$S_1 = \{x \in \mathbb{R}^2: 1 \leqslant \|x \| \leqslant 2 \}$

I want to say that $S_1$ is connected, but not path-connected (since we cannot find a continuous function that takes $0 \in [0,1]$ to $(0,-2) \in S_1$ and $1 \in [0,1]$ to $(0,2) \in S_1).$ How would we demonstrate that it is connected? It seems obvious that we cannot divide $S_1$ into two nonempty disjoint sets since it is a closed annulus, but I am having trouble showing this rigorously.

$S_2 = \{1/n,y): 0 \leqslant y \leqslant 1, n \in \mathbb{N} \} \cup \{(x,0): 0 < x \leqslant 1 \} \cup \{(0,1) \}$

I don't know where to start with this one... any hints would be appreciated

$S_3=\mathbb{R}^2 \setminus \{(x,2):x \in \mathbb{R} \hspace{0.3mm} \} $

I'm pretty sure this is a disconnected space since we have $\mathbb{R}^2$ with entire line $y=2$ removed. $S_3$ is disconnected because we can write $S_3$ as the union of nonempty, disjoint sets: $\{(x,y) \in \mathbb{R}^2: -\infty \leqslant y < 2 \} \cup \{(x,y) \in \mathbb{R}^2: 2 < y \leqslant \infty \}$ Is this argument too hand-wavy?

$S_4= \mathbb{Q} \times \mathbb{Q}$

Is this a disconnected space as well? I'm having trouble visualizing how this would be either a connected or disconnected space if it has irrationals missing from the plane. Hints on this one would also be appreciated.

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  • $\begingroup$ I think it is often useful to think about shapes. Try drawing pictures of $S_1$. I think whether connected/path-connected will become clear. Same goes for $S_3$. At least $S_4$ is a little subtler. Also would you mind clarifying $S_2$? I do not quite see what it is saying. What is the ambient space? $\endgroup$ – user45150 Nov 3 '15 at 3:07
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HINTS: $S_1$ is both connected and path-connected: it’s just a closed annulus. The easiest way to show that it’s connected is to show that it’s path-connected. The easiest way to do that is to observe that it’s the set of points in $\Bbb R^2$ whose polar coordinates have the form $\langle r,\theta\rangle$ for any $\theta$ and $1\le r\le 2$. To get from one point to another you need at worst a radial segment and a circular arc.

Draw a sketch of $S_2$; it’s a sort of comb with one extra point at $\langle 0,1\rangle$. Show that there is no path with $\langle 0,1\rangle$ as an endpoint. Show further that $S_2\setminus\{\langle 0,1\rangle\}$ is path-connected and hence connected; this tells you what the only possible disconnection of $S_2$ could be, and you need only prove that it isn’t one. (Alternatively, use the first part to show that $S_2$ has a dense connected subset and is therefore connected.)

You’re right that $S_3$ is not connected, and you’ve found the right two sets to show this, but it’s not enough to say that they’re non-empty and disjoint: it’s also crucial that their union is $S_3$ and that they are both open in $S_3$.

You can apply the same technique to $S_4$: look, for instance, at the line $y=\sqrt2$.

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  • $\begingroup$ Thanks for the hints. For $S_1,$ would the function just be $f = r \theta$? I think I've figured out the rest. $\endgroup$ – George Dilla Nov 3 '15 at 5:02
  • $\begingroup$ @George: I’m not sure what you mean by $f=r\theta$; I’m suggesting that if the two points are $\langle r_1,\theta_1\rangle$ and $\langle r_2,\theta_2\rangle$, you can send $[0,1/2]$ linearly to the segment from $\langle r_1,\theta_1\rangle$ to $\langle r_2,\theta_1\rangle$ and $[1/2,1]$ to the arc $\langle r_2,\theta_1\rangle$ to $\langle r_2,\theta_2\rangle$. $\endgroup$ – Brian M. Scott Nov 3 '15 at 5:09

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