Is 'smallest set having P' equivalent to 'intersection of all sets having P'?

Question

Are these two equivalent?

1) $S$ is the smallest set having property $P$. That is, $S$ has property $P$, and any set that also has property $P$ is a superset of $S$.

2) $S$ is the intersection of all sets having property $P$.

Intuitively, these seem equivalent. I was easily able to prove $ 1 \rightarrow 2 $, however the reverse seems tricky, and I don't know if it is true. I tried to cook a counterexample, but am unsure if it is correct.

Let $S$ be the intersection of all 1-dimensional subspaces of $\mathbb{R}^{2}$. It can be shown that $S$ is the zero vector space, which is clearly not 1-dimensional. How can I refine definition 2 so that the two are equivalent? I was thinking to add the condition that $S$ has property $P$ to definition 2. Would they then be equivalent? But doesn't this presuppose that the set that is the intersection of all sets having property $P$ also has property $P$?

Answer 1

The problem is that there need not be a smallest set having property $P$. You see this in your example: there is no smallest subspace of dimension 1 of $\mathbb R^2$. On the other hand, there is a smallest subspace without dimension restriction, and indeed you have found it. Thus, if you add the assumption that a smallest set $S$ exists, then you can prove the converse.

In fact, that is usually how you prove a smallest set with a certain property exists. You prove that the intersection of any collection of sets with that property still has that property, and then (almost trivially) the intersection of all of them must be the smallest.

Answer 2

In general, the intersection of even two sets with property P may not have property P. So for your principle to work you would need to assume the property P satisfies: the intersection of an arbitrary family of sets with property P again has property P.

For example, given a set $A \subseteq \mathbb R$ you can get the smallesst closed set containing $A$ using your method. But you cannot get the smallest open set containing $A$, which may not even exist.

Answer 3

I had the same question in mind for sometimes, and I thought we can prove it the same way we do with the principle of induction : first we work on sets and than we associate a proposition (=true) to sets. I'm not a mathematician though, so I could be wrong. In any event here's the steps :

(1) Given a family of sets $\{A_i \}_{i\in I}$, then $\bigcap_{i\in I}A_i\subseteq A_i$, $\forall i\in I$.

proof (1) : let's take an element $x\in \bigcap_{i\in I}A_i$. Then $x\in A_i, \forall i\in I$.

(2) Take a family of sets $\{A_i \}_{i\in I}$ , such that $P(A_i)=true$.
       if $P(\bigcap_{i\in I}A_i)=true$, then $\bigcap_{i\in I}A_i$ is the smallest set of the collection $\{A_i \}_{i\in I}$.

proof (2) : Since we know by ipothesis that $P(\bigcap_{i\in I}A_i)=true$, then we know that $ \bigcap_{i\in I}A_i\in\{A_i \}_{i\in I}$. Then from point (1) we know that the intersection is smaller than any other set in the collection $\{A_i \}_{i\in I}$.

(3) Take a family of sets $\{A_i \}_{i\in I}$ , such that $P(x)=true, \forall x\in A_i$.
       if $P(x)=true, \forall x\in \bigcap_{i\in I}A_i$, then $\bigcap_{i\in I}A_i$ is the smallest set of the collection $\{A_i \}_{i\in I}$.

proof (3) : Since we know by ipothesis that $P(x)=true, \forall x\in \bigcap_{i\in I}A_i$, then we know that $ \bigcap_{i\in I}A_i\in\{A_i \}_{i\in I}$. Then from point (1) we know that the intersection is smaller than any other set in the collection $\{A_i \}_{i\in I}$.