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Question: Show that the set $C=\{(x,y)\in \mathbb{R}^2: 1\leq x^2+y^2<2\}$ is connected.

My Question: My main question is what open sets we should pick in $\mathbb{R}^2$. Once I know what open sets we should pick I will be able to solve this.

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What do you mean by "what open sets we should pick"? –  anonymous Nov 13 '12 at 17:48
    
I wanted to show this is connected by picking any two open disjoint subsets and showing that if C is contained in the union of the two, then C is contained in only one of them. –  Mathstudent Nov 13 '12 at 17:53
    
If you want to go by this definition, then, as you said, you canot choose those open sets. It might be easier to prove that this set is path connected, and then show that this implies connectedness. –  anonymous Nov 13 '12 at 17:58
    
Ok we haven't talked about path connected yet but I will look on my work and see what I can come up with. I will respond when I have seen what I can do with path connected. –  Mathstudent Nov 13 '12 at 17:59
    
One can give an explicit path between any two points $p_1,p_2 \in C$. That is a continuous function $\gamma:[0,1]\to C$ such that $\gamma(0) = p_1$, $\gamma(1) = p_2$. First, note that $\{(x,y) | x^2+y^2 = 1$\} is path connected using a portion of the path $t \mapsto (\cos t, \sin t)$ to connect any two points on the unit circle. Then any point in $C$ can be connected to the unit circle by a straight radial line. This gives a continuous path between any two points in $C$. –  copper.hat Nov 13 '12 at 18:26
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3 Answers

up vote 3 down vote accepted

A standard way to show that $C$ is connected is to observe that it is path connected. Given any two points $A$ and $B$ we can, for instance, move radially from $A$ until we reach a distance from $(0,0)$ equal to that of $B$ and then, if necessary, move on a circular arc to reach $B$.

A standard theorem in general topology tells us that a path connected set is connected.

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This set is path-connected, which is clear if you draw it. It might be a bit tedious but you could actually write down explicitly the path connecting any two points $(s,t)$ and $(u,v)$ by first writing them in polar coordinates, then (if $s^{2}+t^{2}>1$) walk along the ray towards the origin until you reach the circle $S^{1}=\{(x,y);x^{2}+y^{2}=1\}$, rotate along this circle until you reach the correct angle and then again - if needed - move out along the ray coming from the origin towards $(u,v)$.

It is not even needed to move towards $S^{1}$ but this saves you discussion of different cases (the cases mentioned in the post preceding this one). You can w.l.o.g. assume that $t=0$.

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A comment on sdcvvc's answer: actually you could just show that this is the image of a continuous map $f:\mathbb{R}\times[0,1)\to\mathbb{R}^{2}$. The product of connected sets is connected and the image of a connected set under a continuous map is also connected. You do not need that this is actually homeomorphic to $S^{1}\times[0,1)$. –  M. Luethi Nov 13 '12 at 18:10
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Alternatively, prove that $C$ is homeomorphic to $S^1 \times [0,1)$ which is a product of two connected spaces.

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