Inverse Image(Preimage) of Intersection of Sets Proof

Рет қаралды 41,429

The Math Sorcerer

Күн бұрын

Пікірлер: 20

@noisjang 5 жыл бұрын

You've made a beautiful job... Your country must be proud of you!

@oingomoingo7411 4 жыл бұрын

very good video there are very few ppl who explain this with a easy speech pls make more math videos !!

@TheMathSorcerer 4 жыл бұрын

Thank you😀

@TheMathSorcerer 10 жыл бұрын

@Sofialovesmath 2 жыл бұрын

You are such a great teacher

@alexanderzieschang2664 4 жыл бұрын

Why do I have to write it as x∈f(A∩B) first, to show the equality only to change it back again. Couldn't I do the same thing with f^(-1)(A∩B)?

@laykefindley6604 4 жыл бұрын

What lets you just replace intersect with and? Is their a definition that is unmentioned?

@jammcrusader1981 3 жыл бұрын

intersect and, union or - but you prob are over this by now lmao

@laykefindley6604 3 жыл бұрын

@@jammcrusader1981 i know that's what you can do. But what is the law or proof that let's you do this?

@dariustan9951 6 жыл бұрын

Just curious, what's stopping me from substituting your f^-1 with f and f wth f^-1 such that it proves the image of Intersection of sets is the intersection of images of sets?

@heartpiecegaming8932 4 жыл бұрын

Good question! So if you carefully look at the definitions of f(X) and f^{-1}(X) carefully, you can sort of see that the definitions are not symmetric. f(X) is the set of all y such that y=f(x) for some x in X, whereas, f^{-1}(X) is the set of all x such that f(x) in X. And really, the intersection thing follows from the fact that for every x in X, there is EXACTLY one y such that f(x)=y, which need not be the case for "f^{-1}" (as it can happen that for x =/=z, f(x)=f(z)). And it turns out, f(A intersection B)=f(A) intersection f(B) given that f is injective (i.e. there is only one x such that f(x)=y, if such x exists at all). Hope this helps.