Prove the following set is compact

[Theorem.]

Homework Statement

Let K be a nonempty compact set in R²Prove that the following set is compact:
S=\lbrace{p\in R^{2}:\parallel p-q\parallel\leq 1 for some q\in K}\rbrace

Homework Equations

I will apply Heine-Borel- i.e. a set is compact iff it is bounded and closed

The Attempt at a Solution

This is somewhat of a silly question, I know that all I have to do is split the set up into too compact sets A and B (i.e. A+B=S), I then already know how to prove that the sum of two nonempty compact sets is compact. The problem is, I am having trouble splitting the set up ( I am still fairly new to set notation, which doesn't help).

So far I have looked at
A=\lbrace {p:\parallel p \parallel \leq R, p\in R^{2}\rbrace
This set is both closed and bounded and therefore compact. and...
A=\lbrace {q:\parallel q \parallel \leq 1-R, q\in K\rbrace
This set is also bounded and closed.
If I add the two sets I get the right condition through the triangle inequality, but I clearly end up with
A+B=\lbrace {p+q:\parallel p-q \parallel \leq 1 \rbrace
which isn't what I want,
I think more than anything I am just struggling with the set operation, any help would be much appreciated- I have done the hard(er) part of the proof, my understanding is that this part should be simpler, but I am struggling.
Thanks

 

[lanedance]

the definition doesn;t quiet make sense how can |p-q|<q if q is a point in R2?

 

[hgfalling]

What does this mean?

\Vert p - q \Vert \leq q

I mean, isn't q \in K \subset \mathbb R^2?

 

[Theorem.]

Woops sorry guys that was a typo,
should be \Vert p-q\Vert \leq 1

 

[HallsofIvy]

I'm not sure what you mean by "add the two sets". However, to prove S is bounded, you must prove that {||x- y||}, where x and y are any two points in S, is bounded- that is, ||x- y||< A for some number A. If x is in S, there exist p in K such that ||x- p||<= 1. If y is in S, there exist q in K such that ||y- q||<= 1. Further, since K is compact, there exist R such that ||p- q||< R for all p and q in K.

Then ||x- y||= ||x- p+ p- q+ q- y||<= ||x- p||+ ||p- q||+ ||q- y||< 1+ R+ 1= R+ 2.

To show that S is closed, look at the complement of S, the set of all x such that ||x- p||> 1 for all p in K.

 

[Theorem.]

I'm not sure what you mean by "add the two sets". However, to prove S is bounded, you must prove that {||x- y||}, where x and y are any two points in S, is bounded- that is, ||x- y||< A for some number A. If x is in S, there exist p in K such that ||x- p||<= 1. If y is in S, there exist q in K such that ||y- q||<= 1. Further, since K is compact, there exist R such that ||p- q||< R for all p and q in K.

Then ||x- y||= ||x- p+ p- q+ q- y||<= ||x- p||+ ||p- q||+ ||q- y||< 1+ R+ 1= R+ 2.

To show that S is closed, look at the complement of S, the set of all x such that ||x- p||> 1 for all p in K.

That makes a lot more sense Thank you! I am not sure how my professor was going about "splitting" up the set. By adding the two sets, I meant pointwise addition, S_1+S_2 =\lbrace v_1+v_2 : v_1\in S_1, v_2\in S_2}\ In his method he constructed two sets such that they account for all points in S. From there, I proved that sum of two compact sets is compact (Sequentially compact).
Using the method you laid out seems much simpler, and I can't see where it would fail. Thank you for your help : )