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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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    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 $C$.2012-11-13

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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 $S^{1}\times[0,1)$.2012-11-13
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Alternatively, prove that $C$ is homeomorphic to $S^1 \times [0,1)$ which is a product of two connected spaces.