Verification that ZxZ has no cluster points in RxR

[MidgetDwarf]

Homework Statement

ℤxℤ ⊆ ℝ^2 has no cluster points.

Relevant Equations

Definition of neighborhood: If x∈R^n , then any set which contains an open set containing x is called a neighborhood of x in R^n.

Definition of cluster point : x∈R^n is said to be a cluster point of a set A (or a point of accumulation of A) in case every neighborhood of x contains at least one point of A distinct from x.

A set N is a neighborhood of a point x iff there exist an open ball with center x contained in N.

proof:

Assume instead that every point of ℤxℤ is a cluster point. Note that (0,0)∈ℤxℤ. So by assumption, (0,0) is a cluster point of ℤxℤ. ⇒ every neighborhood of (0,0) contains at least one point ℤxℤ different from (0,0).

Consider the open ball with center (0,0) and r =1/2, denoted by A. We know that open balls are open sets ( I proved this before), so A is a neighborhood of (0,0). But A\{(0,0)} ∩ ℤxℤ = ∅. A contradiction.

Therefore, ℤxℤ has no cluster points.


Q.E.D

Not sure if my proof is correct. Moreover, since a set is a subset of itself, and we know that open balls are open sets. Does it follow that an open ball centered at x with radius r is a neighborhood of x? I believe it is, but a part of me is uncertain.

 

[Mark44]

proof:

Assume instead that every point of ℤxℤ is a cluster point. Note that (0,0)∈ℤxℤ. So by assumption, (0,0) is a cluster point of ℤxℤ. ⇒ every neighborhood of (0,0) contains at least one point ℤxℤ different from (0,0).

Consider the open ball with center (0,0) and r =1/2, denoted by A. We know that open balls are open sets ( I proved this before), so A is a neighborhood of (0,0). But A\{(0,0)} ∩ ℤxℤ = ∅. A contradiction.

Therefore, ℤxℤ has no cluster points.

Looks good to me. You can without loss of generality (wlog) work with the point (0,0). The proof will hold for every other lattice point in Z x Z.
You chose r = 1/2, but if you take any r such that r < 1, then no open neighborhood of (0, 0) of radius r will contain another lattice point.

 

[MidgetDwarf]

Ahhh. Thank you for that. Yes, r<1 works. Just for clarification. Should I add the part where you said wlog to my proof? So In general when I am trying to prove either a statement is true or false, say I know that its false. We proceed by contraction, and assume its true for all element in the set. Then we pick one element, and show that it leads to a contradiction. Which proves that the original statement was false?

 

[Mark44]

Just for clarification. Should I add the part where you said wlog to my proof?

"wlog" is a fairly commonly used acronym. If you add that, it says that although you're dealing only with one specific case, you could have picked any case, not just the point (0, 0) in your problem.

So In general when I am trying to prove either a statement is true or false, say I know that its false. We proceed by contraction, and assume its true for all element in the set. Then we pick one element, and show that it leads to a contradiction. Which proves that the original statement was false?

In a proof by contradiction, you're assuming the opposite conclusion, then you look for a counterexample. In your problem, the contradictory statement is that every point in Z x Z has a cluster point. If you can show that for one particular point, say (0, 0), there is not cluster point, then you have contradicted the statement you assumed was true. Therefore that assumption must be false, so the original statement must be true. Is that clear?