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Let A = {(x, -2x): x is in Z}. Show that A is denumerable. Hence show that B = {(-2x, x): x is in N} is denumerable.?
I have already shown that A is denumerable by defining a bijective function from Z to A (since Z is denumerable):
Define f: Z -> A by f(z) = (z, -2z).I can similarly define a bijection from N to B to show that B is denumerable, but I think the question wants me to define a bijective function from A to B. I am not sure what this bijective function could be.I was thinking of a piecewise function where f(x) = -x for x > 0, f(x) = x for x < 0 and f(x) = (x - 1, 2x+ 1) for x = 0. But this seems very complicated since that showing A is denumerable is worth more points than showing that B is denumerable.
2 Answers
- 9 months ago
First, define C = {(x, -2x) : x is in N}. Because N is a subset of Z, C is a subset of A, and because any subset of a denumerable set is also denumerable, C is also denumerable.
Consider the transformation T: C->B such that T(x,y) = (y,x).If two points (x0, y0) and (x1, y1) transform to the same point (y, x), then y0 = y1 = y and x0 = x1 = x, so they are the same point and therefore the function is injective.
Given any point (-2x, x) in B, there is a point (x, -2x) in C such that T maps the point in C to the point in B, so the function is also surjective.
If the function is both injective and surjective, it is a bijection, and because C is denumerable, so is B.
Hope this helps!