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# Preimage of a set

Definition of preimage of a set. Created by Sal Khan.

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• At around 4.45, Sal says that T(T^-1(S)) forms only a subset of S. What confuses me is, if an element in S isn't mapped by the transformation, then that element is not part of S, and is therefore outside S. The preimage of S must, surely, include the preimage of every element in S.
• When Sal defines S he doesn't define it as the result of the transformation, he defines S as just any subset of Y. Therefore this subset S is not limited to those that have a matching element in X.

As a result, there can exist some elements of S that do not have a matching element in the preimage of S and these unmatched elements would therefore not appear in the image of the preimage of S, aka T(T^-1(S)).
• So, just to clarify, the set S is a subset of our CODOMAIN but not necessarily a subset of the IMAGE of our function, right?

And, with regard to his "bonus points" question, the image of our preimage of S under T [ T(T^-1(S)) ] is the intersection of the IMAGE of X under T and S, the subset of our CODOMAIN Y. Correct?
• Yes to the first. There are some elements of S which are not necessarily reached, I believe, by the image of the pre image of S under T.

...and yes to the Second!. Both correct as far as I can see!
• Couldn't there be a bunch of preimages? For example, if the domain is R and the codomain is R and the subset S is all R≥0, then the preimage could be x^2, x^4, e^x... because all of these functions map members of R to members of R≥0
• Actually, nevermind. I realize now that the preimage is a set, not a transformation.
• Hi, is there a proof of this T(T^-1(S)) ⊂ S ?
Thanks
• Let x be an element of T(T^-1(S)). The image T(V) is defined as the set {k | k=T(v) for some v in V}. So x=T(y) where y is an element of T^-1(S). The preimage of S is the set {m | T(m) is in S}. Thus T(y) is in S, so since x=T(y), we have that x is in S.
Thus we have shown if x is in T(T^-1(S)), then x is in S, so T(T^-1(S)) ⊂ S.
• I think that "T^-1(T(A) = A". Am I right?
• Yep! You can think of inverting a transformation as undoing it.
• Is it safe to say that X is a set of n vectors? Same for Y - a set of m vectors?
• Not really. X can be a set of anything. If it were vectors you would NORMALLY describe it as a set of vectors in Rn, so each vector has n elements. Same with Y, vectors would have m elements.

You could techncally have a set with n vectors mapping onto a set with m vectors, but there wouldn' be a lot that could be done with vector arithmetic, or matrix arithmetic.

(1 vote)
• Ok Sal, fair enough, but what if you find preimages recurseively adinfinitum....
T(T^-1(T(T^-1.........
You are finding infinite subsets that may or may not be smaller than the originals?
I'm trying to think of this functionally where f( f^-1(x) ) would not necessarily be x, but some subset of x?
• The problem of what is the co-domain and what is the range of the function comes up again. If T maps to some sub-set of S where S is itself a sub-set of the co-domain, the function defines only this sub-set of S. So how do we define S and the co-domain. They appear to be entirely arbitrary.Any set is a sub-set of the universal set, therefore we can choose anything we like to be the co-domain.
(1 vote)
• Yes. However, in general you won't be working out what the co-domain is. You get a function, and the definition tells you what the co-domain is.. Your logic is right though, and if you were creating a function you might have to work out what you wanted the function to have as a co-domain in order for it to make the most sense in context.. maybe choosing something too far removed from the problem will confuse people you try to describe your function to.