# Problem in real analysis that seems to prove something erroneous

I was working on a real analysis problem that goes as follows: Show that, for an arbitrary function $$g:\mathbb R\to\mathbb R$$ it is always true that $$g(A\cap B)\subseteq g(A)\cap g(B)$$. See, I ended up proving that this statement is true, but that's not what I'm confused about. What I'm confused about is how a proof for the opposite direction is false. I'm about to do some certified $$BS^{tm}$$ and prove that $$g(A) \cap g(B)\subseteq g(A\cap B)$$. Suppose some $$g(x)\in g(A)\cap g(B)$$. Then $$g(x) \in g(A)$$ and $$g(x)\in g(B)$$. It must follow that $$x\in A$$ and $$x\in B$$ by the definition of $$g(A)$$ and $$g(B)$$. So $$x\in A\cap B$$. However, this would imply that $$g(x)\in g(A \cap B)$$ by the definition of $$g(A\cap B)= \{ g(x) \mid x \in A \cap B \}$$. So $$g(A) \cap g(B)\subseteq g(A \cap B)$$.

Of course, this proof clearly has to be wrong, since that would imply the sets are equal (we assume that the correct direction for the subset is proven, although I didn't show the proof here). However, a previous problem showed that there are sets $$A$$ and $$B$$ that make this wrong for certain functions. My question is where my $$BS^{tm}$$ proof of the other direction went wrong. I know the proof shown can't work, but I can't see the flaw.

Edit: apologies for the problematic mathjax, I'm still getting the hang of this.

• Hint: What if $g$ is not injective? (This also stumped me for a good three hours when I was first learning analysis!) Jun 3 at 20:14
• $y\in g(A)\cap g(B)$ does not mean $y=g(x)$ for only one $x.$ Jun 3 at 20:24
• P.S. regarding mathjax you do not need to put  after every symbol, only at the beginning and end of each equation. This is why some of your equations spill into separate lines Jun 3 at 20:25
• "since that would imply the sets are equal" Why is that a problem? Jun 4 at 5:22
• @Acccumulation because there are counter examples where saying those sets are equal doesn't work. The function $g$ only has the property that those sets are equal when $g$ is injective (see GEdgar's comment) Jun 7 at 20:22

No. From $$g(x) \in g(A)$$ we may not conclude $$x \in A$$. This only works if $$g$$ is injective.

Example. Let $$g : \mathbb R \to \mathbb R$$ be defined by $$g(x) = x^2$$. Let $$A= [0,2]$$. Then $$g(A) = [0,4]$$. But $$g(-1) = 1 \in g(A)$$ while $$-1 \not\in A$$.

• Ah I see, thanks for pointing that out! From what @Arturo Magidin said, I assume the essential idea is that $g(x)$ $\in$ $g(A)$ only implies that some $x$ is in $A$, not that specifically the input for $g(x)$ is in $A$, as multiple $x$ could map to the same output in a non-injective function. Jun 3 at 21:09
• @TimothyBennett To be more precise (you are using $x$ for two things there), $g(x)\in g(A)$ implies that there is an $a\in A$ with $g(a)=g(x)$. And this does not tell us that $a=x$ when $g$ is not one-to-one. Jun 3 at 21:27

The error becomes clear when you use notation that is not so (mis)suggestive.

Instead of writing $$g(C)=\{g(x)\mid x\in C\}$$, let us write it as $$g(C)=\{ y\in \mathbb{R}\mid \text{there exists }x\in C\text{ such that } g(x)=y\}.$$ That means that $$y\in g(C)$$ if and only if there exists $$c\in C$$ such that $$g(c)=y$$.

So suppose $$y\in g(A)\cap g(B)$$. Then $$y\in g(A)$$, so there exists $$a\in A$$ such that $$g(a)=y$$. Likewise, $$y\in g(B)$$, so there exists $$b\in B$$ such that $$g(b)=y$$.

Do now we have $$a\in A$$ and $$b\in B$$ with $$g(a)=y=g(b)$$. But we want $$x\in A\cap B$$ with $$g(x)=y$$. And it is now clear that this does not immediately follow from what we have. We would need to somehow conclude $$a=b$$, but without additional assumptions this does not follow. This also tells you how to construct a counterexample, if one is needed: take two disjoint sets that are mapped to the same $$y$$ under $$g$$.