Erdős Problem 702: forbidding singleton intersections
Statement qualification and claim.
For each fixed $k\geq4$, and for all sufficiently large $n$ depending on $k$, a $k$-uniform family on $[n]$ with more than $\binom{n-2}{k-2}$ members contains two sets intersecting in exactly one point. The dependence on sufficiently large $n$ is essential. It appears explicitly in Frankl's primary paper, although absent from the displayed statement at
https://www.erdosproblems.com/702. We state the established theorem as an external input, check the sharp example, and give explicit counterexamples to the unqualified variants.
The known theorem, with its hypotheses retained.
Frankl proved that for each integer $k\geq4$ there is $n_0(k)$ such that whenever $n>n_0(k)$ and $\mathcal F\subseteq\binom{[n]}k$ satisfies \[
|A\cap B|\ne1\qquad(A,B\in\mathcal F,\ A\ne B),
\] one has \[
|\mathcal F|\leq\binom{n-2}{k-2}. \tag{1}
\] This is the principal external theorem used here: P. Frankl,
On families of finite sets no two of which intersect in a singleton, Bull. Austral. Math. Soc. 17 $1977$, 125--134,
https://doi.org/10.1017/S0004972700025521. Taking the contrapositive of $1$ proves the correctly quantified assertion. This use does not purport to supply a new proof of Frankl's structural theorem.
Sharpness of the bound.
Fix two points $a,b\in[n]$ and take \[
\mathcal F=\{A\in\binom{[n]}k:\{a,b\}\subseteq A\}.
\] There are exactly $\binom{n-2}{k-2}$ choices. Every two members meet in at least $a,b$, so none intersect in a singleton. Thus the strict inequality in the question is necessary, and the bound $1$ is sharp.
Why the omitted size qualification cannot be dropped.
For any $k\geq4$, take $n=k+1$ and all $k$-subsets of $[k+1]$. The family has $k+1$ members, whereas \[
\binom{n-2}{k-2}=\binom{k-1}{k-2}=k-1.
\] Two different members intersect in $k-1\geq3$ elements. This disproves the literal assertion for arbitrary $n$; it does not contradict Frankl's theorem.
Why the restriction $k\geq4$ also matters.
For $k=3$ and $n=4t$, partition the ground set into $t$ blocks of size four and take every triple inside each block. There are $4t=n$ triples. Two triples from the same block intersect in two elements, and triples from different blocks are disjoint. Thus no singleton intersection occurs, although \[
n>\binom{n-2}{1}=n-2.
\] This gives arbitrarily large counterexamples when $k=3$. For $k=2$, a matching of two or more edges already exceeds $\binom{n-2}{0}=1$ and has no singleton intersection.
A direct link-family reduction.
For each $x\in[n]$, define $\mathcal F_x=\{A\setminus\{x\}:x\in A\in\mathcal F\}$. The original family has no singleton intersection if and only if every $\mathcal F_x$ is intersecting. Indeed, two disjoint members of $\mathcal F_x$ lift to sets whose intersection is exactly $\{x\}$, and conversely. For $n-1\geq2(k-1)$, the usual Erdős--Ko--Rado theorem on each link gives \[
|\mathcal F_x|\leq\binom{n-2}{k-2},\qquad
k|\mathcal F|\leq n\binom{n-2}{k-2}.
\] This last bound explains why treating the links separately does not recover the sharp result: Frankl's theorem controls their compatibility and removes the extra factor $n/k$.
Scope.
The literal statement without a condition on $n$ is disproved by the explicit example above. The intended large-$n$ statement follows completely from the precisely stated published theorem; its threshold and its two necessary qualifications have been checked directly.
Completion estimate.
COMPLETION: 100%