Why is $g(A) \cap g(B) \subseteq g(A \cap B)$ not true

elementary-set-theory

The following is from Understanding Analysis, Stephen Abbott.

Show that, for an arbitrary function $g : \mathbb{R} → \mathbb{R}$, it is always true
that $$g(A \cap B) \subseteq g(A) \cap g(B)$$ for all sets $A, B
> \subseteq R$.

Question: I want to understand this better by also understanding why the opposite statement is not true: $$g(A) \cap g(B) \subseteq g(A \cap B)$$

My flawed attempt at answering the original question is as follows:

  • let's say $y \in g(A \cap B)$

  • if $x \in A \cap B$ then both of the following statements must be true: (1) $x \in A$ and (2) $x \in B$

  • using only the first statement: if $x \in A$ then $y \in g(A)$

  • using only the second statement: if $x \in B$ then $y \in g(B)$

  • but since both statements have to be true, we can say $y \in g(A) \cap g(B)$

  • that is, $g(A \cap B) \subseteq g(A) \cap g(B)$

The reason I believe this is flawed, is because it doesn't allow me to say why the opposite $g(A) \cap g(B) \subseteq g(A \cap B)$ is not true.

Exploration

I can think of examples to show why the opposite is not true, for example:

  • $g(x) = x^2$ and $A=[-2,-1], B=[1,2]$ has $A \cap B = \emptyset$, but $g(A) = g(B) = [1,4]$

One of the ideas I am trying to explore is which step of the above proof would not work for the opposite:

  • I can "split the domain", a statement about $A \cap B$ into two statements, one about $A$ and one about $B$.
  • But I can't "merge the co-domains", statements about $g(A)$ and $g(B)$ into a single statement about $A \cap B$
  • And the reason for this is because the co-domains might not be 1-to-1 unique (injective function?).

I would appreciate identifying the flaw in the attempted proof, which will I hope shed light on why the opposite statement is not true.

Best Answer

Great counterexample, I would've used the same one! It doesn't seem like it helps us, but it does. In your counterexample, think about how 1 and -1 use the fact that a function is not by definition one-to-one (injective). Or that $g(a)=g(b)$ does not necessarily imply $a=b$.

This method will not help us the opposite way (the one it seems you were originally given) because a function IS by definition, one to one. I.e., $a=b$ DOES necessarily imply that $g(a)=g(b)$. So every $y\in g(A\cap B)$ has an $x$ with $x\in A$ AND $x\in B$ so that $g(x)=y$. Therefore $g(x)=y\in g(A)$ AND $g(x)=y\in g(B)$, so we have $g(x)=y\in g(A)\cap g(B)$.

Or you may think of it that the $g(A)\cap g(B)$ set is "bigger" because you don't intersect $A$ and $B$ first, so:

  • If g is injective, then the sets $g(A\cap B)$ and $g(A)\cap g(B)$ are the same size.
  • If g is not injective, then more elements may be in $g(A)\cap g(B)$ because of elements that are not common to both $A$ and $B$ but are common to both $g(A)$ and $g(B)$.

On the other hand, if we intersect $A$ and $B$ first, the new set's image over $g$ can't be bigger than intersecting the images because if our starting real number is in $A$ AND $B$, then its image is definitely in each of the individual images.

This was my first stack exchange answer, I hope it was clear. I tried to lean on function-based thinking more than set theory thinking because most of us learn functions first and might be more comfortable with it.

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