Design an algorithm to generate all possible (distinct) subsets of a given set (finite) of distinct elements.

Answer 1

WHY ???????????

Answer 2

to assert that contraptions do no longer contain duplicate factors is somewhat deceptive. you have gotten a sequence of balls such as 10 white balls and 12 black ones. you have a sequence of twenty-two gadgets. probability questions are rife with operations on such contraptions. how many procedures are you able to create a subset? for each ingredient, it may the two be interior the set or no longer interior the set. There are 2 opportunities for each merchandise. Multiply that out, and you will see the two^n is the attainable style of subsets, as many have already mentioned. it relatively is purely yet another handle it. you are able to characterize each and each subset in binary notation, with n binary digits. affiliate between the digits with each and each merchandise, and set it = one million if that's interior the subset, 0 if no longer. this supplies yet another rapid thank you to count quantity the subsets. E.g., for a sequence of four gadgets, the numbers run from 0 (or 0000) to 1111 (binary notation for 15. 0 to fifteen is sixteen numbers, = 2^4 0 is the empty set, and 1111 is the entire set, the two considered one of that are seen subsets of the entire set. A subset does not must be a appropriate subset, nor non-empty. purely yet in a different thank you to look at it.

Answer 3

If you are seriously worried by this question, then you should think about what can possibly be expected from any attempt to provide a foundation for mathematics. As Lewis Carroll noted, if I ever try to demonstrate that some reasoning is sound, then you can always ask why the demonstration itself is sound. All justification has to come to an end somewhere (and now I am quoting Wittgenstein out of context). There is a further, very serious obstacle, which is that, as Gödel showed, it is impossible to prove the consistency of ZF within ZF. This means that we have to take the consistency of our axioms on faith, or else adopt the pragmatic attitude that there doesn't seem to be any short proof that they are inconsistent, and if anybody does manage to find an inconsistency, then somebody else will probably manage to find a new set of axioms that avoids the inconsistency and still allows us to do all normal mathematics.

Given these difficulties, what can we say about the great results of logic, such as the independence of the axiom of choice and the continuum hypothesis? The rough structure of such a result is as follows. To prove, for example, that the axiom of choice is unprovable in ZF, one assumes that a model of ZF exists (i.e., that ZF is consistent) and builds out of this model a new model in which the axiom of choice is false. The process of building uses the axioms of ZF. This shows that if ZF+(not AC) is inconsistent, then so is ZF. Of course, the reasoning might be invalid because ZF was not to be trusted, but that does not affect the conditional statement: if you are happy with ZF (from the point of view of consistency) then you are just as happy with ZF+(not AC).

It is fair to say that, amongst professional mathematicians, there is much less interest in the foundations of mathematics than there was a hundred years ago. I myself do not feel particularly reassured to know that the integers can be constructed out of sets - I am happy to take the integers, with their familiar properties, as given, and build up from there. This sort of attitude is fine for most purposes, but occasionally problems emerge from `normal' mathematics which turn out to be statements independent of ZFC. To establish the independence, it is of course important to understand in detail about how to construct models of set theory (in a way that, as I hope is not too painfully obvious, I do not).

Answer 4

Think of it this way:
Sets containing one element
Sets containing two elements
Sets containing three elements
Sets containing N elements

Now how do you get these sets? {A,B,C,D} nice small set. It's subsets are.
{A}
{B}
{C}
{D}
{A, B}
{A,C}
{A,D}
{second element, third element}
{third, fourth}
and of course EMPTY SET.

Answer 5

Not exactly a simple question, but :

http://www.math.nyu.edu/~pach/publications/Preprint.pdf

Answer 6

Do your own school work.

Answer 7

Okay, done that one - what's the next question?

Answer 8

Okay. Done it. Now what?

Answer 9

do it yourself, im a little busy!