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.

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    $\begingroup$ Hint: What if $g$ is not injective? (This also stumped me for a good three hours when I was first learning analysis!) $\endgroup$
    – Andrew
    Jun 3 at 20:14
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    $\begingroup$ $y\in g(A)\cap g(B)$ does not mean $y=g(x)$ for only one $x.$ $\endgroup$ Jun 3 at 20:24
  • $\begingroup$ 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 $\endgroup$
    – duelspace
    Jun 3 at 20:25
  • $\begingroup$ "since that would imply the sets are equal" Why is that a problem? $\endgroup$ Jun 4 at 5:22
  • $\begingroup$ @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) $\endgroup$ Jun 7 at 20:22

2 Answers 2


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 $-1 \not\in A$.

  • $\begingroup$ 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. $\endgroup$ Jun 3 at 21:09
  • $\begingroup$ @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. $\endgroup$ 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 $g$.


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