Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $A$ and $B$ are subsets of a topological space and $f$ is any function from $X$ to another topological space $Y$. Do we have always $f(A \cap B) = f(A) \cap f(B)$?

Thanks in advance

share|improve this question
8  
As stated, the topology is irrelevant: this is true for all $A$, $B$ if and only if $f$ is injective. Did you try any examples? –  Dylan Moreland Jul 14 '12 at 13:14
add comment

5 Answers

up vote 5 down vote accepted

Let $y \in f(A\cap B)$. So there is an $x \in A\cap B$, so $f(x) = y \in f(A\cap B)$. Then obviously $x \in A$, so $y = f(x) \in f(A)$. Also $x \in B$, so $y = f(x) \in f(B)$. This proves that $f(A\cap B) \subseteq f(A) \cap f(B)$.

Now for the other way: as an example, say that $f: \mathbb{R} \to \mathbb{R}$ and $A = [0,1]$ and $B = [2,3]$, can you find both sides for a simple example of $f$?

share|improve this answer
add comment

I find a Venn diagram helpful.

venn

The set $f(A\cap B)$ is the purple on the RHS, whereas $f(A)$ is the union of purple and red and $f(B)$ is the union of purple and blue on the right. In order for $f(A\cap B)=f(A)\cap f(B)$, the red and blue regions in the codomain (the right) must be disjoint. However, it should be clear that, at least with sets, this is not generally the case, because all three regions on the right can be controlled for independently by controlling the function $f$ as one wishes. (This aid also helps with OJ's exercise.)

share|improve this answer
add comment

Consider the case where $f$ is constant and $A$ and $B$ are disjoint.

share|improve this answer
add comment

No, they are not equal in general, but you might be able to prove that one side is a subset of the other. Try it out.

Added later: I would also recommend trying to see what you can prove if you look at unions instead of intersections ... and then see what you can find out if you use $f^{-1}$ instead of $f$ - mathematics can be (even) more fun when you try to find things out for yourself!

share|improve this answer
add comment

enter image description here

Here, $f: A \rightarrow B$ is in green and $\{S_i\} = S_i$ for $S = A, B$ and $i = 1, 2.$
This picture proves that $ f(A_1) \cap f(A_2) \neq f(A_1 \cap A_2)$. Incidentally, the same picture works for Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality.

share|improve this answer
    
What is $S_i$ and what do you mean by $\{S_i\}=S_i$? This is impossible for sets given their usual definition because of the axiom of foundation. –  Daniel Rust Feb 18 at 12:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.