Are there an infinite set of sets that only have one element in common with each other?

This is not a complete explanation, but a summing up of my observations in the comments above.

Consider the projective plane ${\Bbb P}^2({\Bbb F}_q)$ over the field with $q$ elements ($q$ must be of the form $q=p^f$ for some prime $p$). Then the following facts follow more or less trivially from the definitions:

• ${\Bbb P}^2({\Bbb F}_q)$ consists of $1+q+q^2$ points;

• ${\Bbb P}^2({\Bbb F}_q)$ contains $1+q+q^2$ lines, each of them containing $q+1$ points;

• every two lines meet at a single point;

• every point is contained in exactly $1+q$ lines.

Thus, if we call "symbols" the points, and "cards" the lines we have a situation which is exactly thatdescribed in the question.

The problem is that the numeric data do not correspond: if we take $q=7$ so to match the $8$ symbols in each card, the number of cards and of symbols should be $1+7+7^2=57$.

Then, either you lost 2 cards [ :-) ], or I'm left clueless.

You can always define, for a set $I$ (which will be the index set, finite or not), something like this : Define $S_i = \{ \{i,j\} \, | \, j \in I \} $. (Since I chose the two-element subset $\{i,j\}$ instead of the ordered couple, order doesn't matter here.) Clearly $S_i \cap S_j = \{\{i,j\}\}$ for $i \neq j$. But the sets $S_i$ are "just as big" as $I$ itself.

For instance, if $I = \{1,2,3,4\}$, then $S_1 = \{ \{1,1\} , \{1,2\}, \{1,3\}, \{1,4\} \}$, $S_2 = \{ \{2,1\} , \{ 2, 2\} , \{2,3\}, \{2,4\} \}$ and so $S_1 \cap S_2 = \{ \{1,2\} \}$.

Would that be a good example? (The way I came up with this is : imagine for some point in $\mathbb R^2$ on the $x=y$ line that some line goes vertical through this point and another one horizontally. Then for two points $(x_0,x_0)$ and $(x_1, x_1)$ on the $x=y$ line, the intersection of the vertical/horizontal lines of the first point and the vertical/horizontal lines of the second point are precisely $(x_0,x_1)$ and $(x_1,x_0)$. By choosing "non-ordered couples" I get unicity. You can produce an example with some large number of elements in common by simply choosing more "components" and considering ordered tuples instead.)