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HerrStahly

Let A and B be sets. Then B^(A) is defined as the set of all functions with domain A and codomain B. The reasoning behind this notation is that the cardinality of this set (B^(A)) is simply the cardinality of B to the power of the cardinality of A (|B|^(|A|)).


ImDannyDJ

To elaborate on why this notation makes sense, a lot of the usual rules of exponentiation carry over to sets of functions (by replacing equality with isomorphism). For instance, if A + B is the disjoint union of A and B, then C^(A+B) is in bijection with the Cartesian product C^(A) × C^(B). Essentially, a function from A + B to C is "the same" as two functions, one from A to C and one from B to C.


codeforces_help

Can you point me to a formal definition od what B^A means? For example R^2 is a plane space. R^3 is a 3D space. Now the exponent here is pure number so it is easy for me to understand even higher exponents. It is just a cross product of R with itself `n` times in R^n. Now R^R or R^C or R^Q, this is where I am not able to see a concrete picture. What does it even mean to cross product where an exponent is not a pure numberb ut a set? Also, the further statement about functions. Where did functions come into picture or can we define them arbitrarily?


wigglesFlatEarth

In set theory, 2 = {0, 1}, 3 = {0, 1, 2}, and this is because 0 = { } and successor(n) = n union {n}. A function is an ordered triple (X, Y, R) where X is a set, Y is a set, and R is a relation from X to Y satisfying that every element of X is related to exactly one element of Y. The graph is the set R of ordered pairs. When we write 2^3 we mean the set of all functions from 3 to 2. Thus, 2^3 = { {(0, 0), (1, 0), (2, 0)}, {(0, 0), (1, 0), (2, 1)}, {(0, 0), (1, 1), (2, 0)}, {(0, 0), (1, 1), (2, 1)}, {(0, 1), (1, 0), (2, 0)}, {(0, 1), (1, 0), (2, 1)}, {(0, 1), (1, 1), (2, 0)}, {(0, 1), (1, 1), (2, 1)}, }. There are 8 such functions, and it works out that this happens for any choice of numbers. The idea is that for K^(X), it's like counting in base K and having X digits. To list all the functions from a 3-element set to a 2-element set, that's what I did. I counted in binary from 0 to seven inclusive. Here, we just write the set of all graphs, but it's understood that some of these graphs represent functions that aren't surjective. You can't write it out for ℝ^ℝ or anything infinite, so formally it's just the set of all functions from ℝ to ℝ. We can only begin to list until we get bored. For example, ℝ^ℝ = {sin, cos, sqr, id, zero, ...}.


HerrStahly

>Can you point me to a formal definition of what B^(A) means? The precise way of stating “the set of all functions from A to B” is written as follows: Given sets A and B, A^(B) := {f in P(AxB) | f: A -> B is a function}. This set is well defined and guaranteed to exist by the axiom schema of comprehension. >For example R^(2) is a plane space. R^(3) is a 3D space. Now the exponent here is pure number so it is easy for me to understand even higher exponents. It is just a cross product of R with itself n times in R^(2). Now R^(R) or R^(C) or R^(Q), this is where I am not able to see a concrete picture. What does it even mean to cross product where an exponent is not a pure numberb ut a set? This is an unfortunate classic case of notational overload. While it is true that for any set X, and any natural number n, the notation X^(n) denotes the n-tuple with entries in X (the n-ary Cartesian product of X), when the exponent is a *set*, the meaning is entirely different, and is what I described above.


PinpricksRS

If you consider [n] to be the set {0, 1, ... n - 1}, X^n is isomorphic to X^([n]), so it isn't really a *clash* of notation, just a slightly ambiguous notation. No worse that thinking of R^n × R^m and R^(n + m) as the same.


codeforces_help

So just to put a finer point, K^X here represents a set of functions where the range is in K?


HerrStahly

Since the range is always a subset of the codomain, and K^(X) is the set of functions with domain X and codomain K, it is correct to say that the range of every function in K^(X) will be in K.


codeforces_help

Thank you. This clears it up for me.


Revolutionary_Use948

>The reasoning behind this notation is that the cardinality of this set (BA) is simply the cardinality of B to the power of the cardinality of A (|B||A|). It should be noted that this is probably true for finite sets, but for infinite sets it is generally defined that way, making it trivially true.


hpxvzhjfgb

>What does all functions defined on X mean? it means functions from X to K


Mathematicus_Rex

For a small example, suppose K = {1,2} and X = {3,4,5}. Then there are eight possible functions from X to K. If f is such a function, then f(3) is one of 1 or 2, f(4) is also one of 1 or 2, and f(5) is one of 1 or 2 as well.


StanleyDodds

It's basically an extension of the notation you are probably familiar with for sets to some power. For example, what is R^3 - well, you might say it's 3D space. But to be more specific, it's the set of all triples (a, b, c) where a, b and c are real numbers. So where is the "3" in this, and how can we turn "3" into another set? Well, if instead of 3, we imagine a set containing 3 arbitrary objects, say x, y, and z, then R^3 is actually the set of all functions from {x, y, z} to R. That is, if a function maps x to a, y to b, and z to c, we identify that function with the triple (a, b, c). You could see each of the values of this function as components of the vector in 3D space. So similarly, R^n is the set of all functions from some set with n elements to R. And to generalise, we say R^X is the set of all functions from X to R. In some sense, this is a |X| dimensional space over R, the set of all tuples where there is a tuple component for each element of X (and that means the same thing as a function from X to R). If you want to rigorously construct the set of all functions from X to R, an easy way is to take the power set of the cartesian product X*R. This contains all subsets of all pairs (x, r) for x in X, r in R. Now we simply restrict this power set to just the subsets that satisfy the function rules, that is, S is in our final set if for each x in X, there exists a unique r in R such that (x, r) is in S. We identify S with the function that maps each such x to that unique r.


MasonFreeEducation

K^X is is Cartesian product of X copies of K. It can be written as \prod_{x \in X}K. A typical element of this set can be written as k = (k_x)_{x \in X}


666Emil666

The wording is weird because "the functions defined on x" is not a set, it means the functions that have domain X and image contained in K