Why cant the proof for Image of Union applicable to Image of Intersection? I'm new here, and I hope you guys don't mind a simple question.  I'm at my wit's end, and I can't seem to figure this out.  I hope it's ok to post this here.  Let me start with the following givens:
S and T are sets.  Let R⊆S×T be a relation.  Let S1 and S2 be subsets of S.
I am struggling to understand why the proof for this:
R[S1] ∪ R[S2] ⊆ R [S1 ∪ S2]

Cannot be applied to this:
R[S1] ∩ R[S2] ⊆ R [S1 ∩ S2]

the proof can be seen here:
https://proofwiki.org/wiki/Image_of_Union
I understand that there is a counterexample to the intersection of images that shows, for the most part (bijective relations notwithstanding), R[S1] ∩ R[S2] ⊆ R [S1 ∩ S2] does not usually hold.  However, it dawned on me when doing the proof for the union of images that if I followed the same template, I would show that indeed R[S1] ∩ R[S2] ⊆ R[S1 ∩ S2] should hold--even though I know counter examples exist to show it shouldn't.  I am missing something logically here, and it has been eluding me for the past day.  Could someone explain what I'm missing here?  Here is a brief sketch of this proof:
Suppose t ∈ R[S1] ∩ R[S2] =>
t ∈ R[S1] and t ∈ R[S2] =>
∃s : s ∈ S1 and s ∈ S2 : t ∈ R[s]   
∃s ∈ S1 ∩ S2: t ∈ R[s]
t ∈ R[S1 ∩ S2]
 A: Yes, you're missing something here.
The first step of the "alleged" proof is correct:
Suppose $t\in R[S_1]\cap R[S_2]$. This implies that $t\in R[S_1]$ and $t\in R[S_2]$. 
However, and this is where your attempted proof goes wrong, this does not imply that there is one element that is a member of both $S_1$ and $S_2$ such that its image under $R$ is $t$. What that implies is that there are two elements $s_1$ and $s_2$ (and possibly $s_1\neq s_2$) s.t. $s_1\in S_1$ and $s_2 \in S_2$ and $R(s_1)=t=R(s_2)$.
Hope this helps!
A: A useful approach when you are having trouble determining why your "proof" of a false statement does not work is to try it out on a counterexample. What you're looking for is the first false statement $S$. Then the implication from the previous statement $P$ to $S$ is necessarily invalid.
Consider the relation $R=\{(0,0),(0,1),(1,1)\}$ on the the set $\{0,1\}$. We know that $1\in R[0]\cap R[1]$ while $1\notin R\left[\{0\}\cap\{1\}\right]$. So, let's consider the case $t=1,S_1=\{0\},S_2=\{1\}$. The first statement (line) is true. The second one is true as well. Next, let's look at the third line. It fails quite badly. The part $\exists s,s\in S_1\text{ and }s\in S_2$ is already false, since the intersection $S_1\cap S_2$ is empty in our case.
We conclude that it's your second step that is false. Namely $t\in R[S_1]$ and $t\in R[S_2]$ does not imply the existence of $s$ such that $s$ belongs to both $S_1$ and $S_2$ and $t\in R[s]$. The correct statement is "$\exists s_1\in S_1, t\in R[s_1]$ and $\exists s_2\in S_2,t\in R[s_2]$" from which you cannot draw the conclusion that $t\in R[S_1\cap S_2]$.
