Proving Compactness of K in BR(0)

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Homework Statement



R>0, let K be a closed subset of C such that K \subset BR(0) (so K is compact). Show that there exists 0 < r < R such that K\subset Br(0).

Homework Equations


The Attempt at a Solution



Can I write BR(0) = {x\inC : d(x,0) \leqR}?
I know that a compact set is closed and bounded.
Is it something to do with us using \subset and not \subseteq?
As if it was \subseteq then maybe K = BR(0) then there wouldn't be an r. But as K is strictly contained in the ball there must be a bit of room for manoeuvre.

These are my thoughts on this problem. I'm not sure if they're correct or what the question is asking, and if they are I don't know how to write them more formally?

Thank you :)
 
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Mathsgirl said:

Homework Statement



R>0, let K be a closed subset of C such that K \subset BR(0) (so K is compact). Show that there exists 0 < r < R such that K\subset Br(0).

Homework Equations





The Attempt at a Solution



Can I write BR(0) = {x\inC : d(x,0) \leqR}?
I know that a compact set is closed and bounded.
Is it something to do with us using \subset and not \subseteq?
As if it was \subseteq then maybe K = BR(0) then there wouldn't be an r. But as K is strictly contained in the ball there must be a bit of room for manoeuvre.

These are my thoughts on this problem. I'm not sure if they're correct or what the question is asking, and if they are I don't know how to write them more formally?

Thank you :)

From the context of the problem, they must mean B_R(0) to be the open ball, d(x,0)<R. Not the closed ball. Otherwise, it wouldn't be true.
 
Oh that makes more sense.

I think I need to say something like you can always fit a smaller ball inside an open ball.

I remember in real analysis I showed for a non-empty bounded above set E and epsilon >0, there exists an x in E such that supE - epsilon < x \leq sup E. Is it something like this?
 
Mathsgirl said:
Oh that makes more sense.

I think I need to say something like you can always fit a smaller ball inside an open ball.

I remember in real analysis I showed for a non-empty bounded above set E and epsilon >0, there exists an x in E such that supE - epsilon < x \leq sup E. Is it something like this?

Something 'like' that could be made to work. But it's a lot easier if you use the definition of 'compact' and a covering of K by open balls.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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