Let Y = X union {u,v} .

                The image of F restricted to Y^2 contians n, so 
                it remains only to show that F restricted to Y^2 is injective.

                Y^2 = X ^2 union X x { u } union X x {v} union {u} x X union {v} x X union 
                          {u,v}^2.

                F is injective on each of the pair-wise disjoint parts of Y^2 unioned,
                so it will suffice to show that the images of F on each of these 
                are pair-wise disjoint.
		
	    if s is in the image of F on X^2,         |s -   0| < 3 N.
                if s is in the image of F on X x { u },  |s - 2u| < N.
                if s is in the image of F on X x { v},   |s - 2v| < N.
                if s is in the image of F on {u} x X,    |s -   u| < 2 N.
                if s is in the image of F on {v} x X,    |s -   v| < 2 N.
                if s is in the image of F on {u,v}^2,
                    s is in { 3 u, 3 v, u + 2 v, v + 2 u }.

	  The fact that numbers  0, 2u, 2v, u, v, 3u, 3v, are all spaced at least 10 N apart                                   
              shows that the images of F on X^2, Xx{u}, Xx{v}, {u}xX, {v}xX, and the set
              {3 u, 3 v}, are all pair-wise disjoint. 
	  Let A be the union of these, i.e. A = image F on X^2 union image F on Xx{u} ... union {3 u, 3 v}

	  It remains to show that { u + 2 v, v + 2 u } is disjoint from A.

               We had u + 2 v = n by construction.   |n - 0| < N gives |u + 2 v - 0 | < N,
               and by assumption n was not im the image of F on X^2, so 
               u + 2 v is not in A.

	  Let s = v + 2 u.
             Then s - (-3v) = 2 u + 4 v = 2 n, so that | s - (- 3v)| = | 2 n | < 2 N
             | u + 2 v | < N gives |3/2 u - ( -3v )| < 3/2 N
             Combining, |s - 3/2 u| < 3/2 N + 2 N < 4 N
             But 3/2 u is at least 5 N distant from 0, u, v, 2 u, 2 v, 3 u, and 3 v,
             so that s cannot be in A, and cannot be u + 2 v = n.

             This completes proof of the Lemma.

************************

To prove the claim, let Kn be an enumeration of the integers.
There exists a sequence of sets Bn having the properties
1)  Bn+1 contains Bn
2)  F is injective on Bn^2
3)  The image of F on Bn^2 contains Kj for all j <= n

For let B1 be the set { 0, K1 }.  
Given Bn, if Bn already contains Kn+1 let Bn+1 = Bn.
Ohterwise, by the Lemma there exists Bn+1 containing Bn such that
F is injective on Bn+1^2 and the image of F on Bn+1^2 contains Kn+1.

Now let  X be the union of all of the Bn.
Any two elements of X^2 are both contained in Bn^2 for some n, thus
F is injective on X^2.  Given any integer, there is some Kn equal to it,
which is then in the image of F on Bn^2, and so in the image of F on X^2,
making F surjective onto the integers.

This completes the proof of the claim.