(a) If x, y ? R, and x<y, then there is t ? Q such that x<t<y. (b) For a subset

Question

(a) If x, y?R, and x<y, then there is t?Q such that x<t<y.

(b) For a subset A?R, and for an element x?R,we will write x+A for the set

{x + y | y ? A}.

(c) A family of sets, F is called pairwise disjoint if for all A, B ? F, either A = B or

A ? B = ?. (I.e. any two sets in the family are either equal or disjoint.)

Here are the statements to prove:

(1) For any real numbers x and y, the sets x + Q and y + Q are either equal or disjoint.

(2) For each real number x, the set (x + Q) ? (0, 1) is nonempty.

(3) Let F = {(x + Q) ? (0, 1)| x ? R}. Then F is a pairwise disjoint family of sets.

Let E be a set that contains exactly one element from each set in F. It may seem obvious that such a set exists

Explanation / Answer

(1) If both x and y belongs to Q, then x+Q = Q = y+Q, then they are same

If either of x or y does not belong to Q, then they are disjoint since one is set of irrational and the other is rational

If both x and y do not belong to Q, then they are same only if x=y otherwise, they are disjoint

(2) True, since for any x, take a to be fractional part of x and b to be decimal part of x ( i.e if x = 5.677, then take a = .677, b=5 )

Then clearly 0<(x-b)<1 . Suppose fractional part if zero, then x is surely an element of Q and then x+Q=Q intersection (0,1) is non empty. Thus, there intersection is non empty

(3) If x belongs to Q, then Q intersection (0,1) consists of all rational number between 0 and 1

And for each x in RQ ( irrational number ) , x+Q forms a distinct subset of rational number.

So, F is pairwise disjoint.

i.e if x and y are two distinct irrational number then x+z ot equal to y+w for any z and w in (0,1)

(4) q ot equal to r => q+E and r+E are disjoint

If z belongs to both q+E and r+E, then z = q+e = r + e' for some e,e' in E

=> z-e and z-e' are rational, but atleast two of z,e,e' are irrational so, e=e' ( two distinct rational can not make same irrational into rational by substraction)

Thus, q=r ( we get contradiction

(5) We already saw that for x in Q, we get one set and for each irrational we get distinct set .

So, U{q+E |q in Q} contains all irrational number by putting q=0 and all irrational number by taking Q. Thus, it is equal to R

(6) I am unable to read the questions

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