prove or disprove: for every nonempty set A there exists an injective function f: A - P(A).
hint: observe that for each a element of A, the set {a} is an element of P(A).
2006-11-29 15:14:35 · 4 answers · asked by st234 2 in Science & Mathematics ➔ Mathematics
i take it that p(a) is the power set. So that is equal to all possible combinations of set A.
The cardnality of p(a) is also always 2^n where n is the number of elements of set A. You can prove that through induction.
If f : X → Y is an injective function, then Y has at least as many elements as X, in the sense of cardinal numbers.
given that the cardnality of p(a) is always strictly larger than A (even for the empty set) we can say that all elements of set A have an injective relation with p(a). They are not strickly bijective (one-to-one). Here is a nice picture of injective of what it would look like: http://upload.wikimedia.org/wikipedia/commons/thumb/0/02/Injection.svg/200px-Injection.svg.png
also the converse.
P(a)→ A is always surjective.
2006-11-29 15:30:19 · answer #1 · answered by xian gaon 2 · 0⤊ 1⤋
No... math can not teach or disprove God, yet there's a complicated gadget of numbers woven into the scriptures which replace into utilized via the Pharisees to examine the accuracy of all the hand-written Torah manuscripts. the comparable quantity varieties additionally happen in the hot testomony Bible scriptures I definitely have seen a quantity code this is derived from a clue in Revelations... It supplies a potential of figuring out the guy who may be the antichrist ... some years in the past I hunted for extra data that this code replace into no longer purely a accident. during the subsequent couple of weeks I had chanced on a itemizing of words that extra as much as the comparable quantity & each a variety of words had a non secular connotation...They have been all against the Lord, or as an adjective - they have been "anti-Christ". He even advised me one among them! after i could been examining this for hours I had positioned it aside to pass to sleep and a voice advised me to "attempt - -- -" (a recognition which additionally extra to that quantity) The staggering element replace into that this education have been given in Greek and historical Hebrew... purely God knew back then that English may be the main language of our time.
2016-12-29 16:52:37 · answer #2 · answered by Anonymous · 0⤊ 0⤋
If by P(A) you mean the power set of A, then it's false, since the power set always has more elements.
2006-11-29 15:23:41 · answer #3 · answered by bictor717 3 · 0⤊ 1⤋
So I guess the function f(x) = {x} works fine?
2006-11-29 15:20:35 · answer #4 · answered by stephen m 4 · 0⤊ 0⤋