Question

\[let \ f:\mathbb{R}\to \mathbb{R} \ and \ one \ to \ one \ and \ A,B\subseteq \mathbb{R}. \ Prove, \ f(A \cap B)=f(A)\cap f(B)\]

Answer 1

@Kainui

Answer 2

@Loser66 canyou figure out this?

Answer 3

Real analysis? haizzz, I am lazy to look up the note

Answer 4

ok, let me try

Answer 5

Thank you

Answer 6

I've done work on this question. Would you like to see it?

Answer 7

rather "some work"

Answer 8

I first assumed that f(A intersection B) does not equal to f(A) intersection f(B)

Answer 9

(I tried to do it using contradiction method.)

Answer 10

That means there exists some x belong(s) to A intersection B such that f(x) does not belong to f(A) intersection f(B)

Answer 11

or

Answer 12

Let \(y\in f(A\cap B)\) then there exist \(x \in (A\cap B)\) such that f(x) = y
\(x \in (A\cap B)\), hence \(x\in A\) and \(x\in B\)

Answer 13

\(x\in A\) then \(f(x) \in f(A)\)
\(x\in B\) then\(f(x)\in f(B)\)
-----------------------------
\(f(x) \in f(A)\cap f(B)\)
hence \(f(A\cap B) \subset f(A)\cap f(B)\)

Answer 14

one way is done, now another way

Answer 15

ok...

Answer 16

just go back ward.

Answer 17

you know that we need prove both ways, right?