# Math Help - real analysis question

1. ## real analysis question

i need to determine and prove whether or not the annulus in R^2 defined by d((x_1,x_2),(y_1,y_2))=sqrt((x_1-y_1)^2+(x_2-y_2)^2) is path-connected.

i know that it is path-connected (logically) but i got stuck at the part where we need to find a path between two arbitrary points in the annulus. can someone help me? thanks.

2. Originally Posted by squarerootof2
i need to determine and prove whether or not the annulus in R^2 defined by d((x_1,x_2),(y_1,y_2))=sqrt((x_1-y_1)^2+(x_2-y_2)^2) is path-connected.

i know that it is path-connected (logically) but i got stuck at the part where we need to find a path between two arbitrary points in the annulus. can someone help me? thanks.
Lets assume you mean the region $R=\{(x,y), 1/4\le x^2+y^2 \le 1\}$.

Take any two points, the line segment conecting them lies either entirly within R or it meets the inner boundary at two points. In the latter case connect the two ponits on the inner boundary by the minor arc.

Thus either two points are connected by the line segment between them if this is within the annulus or by a pair of line segments and an arc otherwise.

Alternativly take the two points as $(r_1, \theta_1)$, $(r_2, \theta_2)$ in polars, then $1/2 \le r_1 \le 1$ and $1/2 \le r_2 \le 1$.

Then the curve:

$
C = \{ (r_1 ,\theta ),\quad \theta \in [\min (\theta _1 ,\theta _2 ),\max (\theta _1 ,\theta _2 )]\} \cup \{ (r_1 + \lambda (r_2 - r_1 ),\theta _2 ),\quad \lambda \in [0,1]\}
$

lies entirly within the annulus and conects the two points

Modifying these to apply to any annulus I leave to you.

RonL

3. hmm can you please explain how that curve works? i'm trying to figure it out and i'm not quite getting the picture. thanks.

4. Originally Posted by squarerootof2
hmm can you please explain how that curve works? i'm trying to figure it out and i'm not quite getting the picture. thanks.
It's an arc of a circle starting from one of the points to the polar angle of the pther point, then proceeds radialy out to the the second point.

RonL