1

If an element is contained in a set, and this set is contained in another set, does the second set contain the element? In ordinary language I'd say yes, but my mathematical intuition says no. What's the right answers according to set theory?

amWhy
  • 210,739
Adrian
  • 1,737
  • 2
  • 17
  • 22
  • see also this – user 1 Jan 14 '15 at 17:01
  • 1
    If it helps, don't use "is contained in" when you mean "is an element of" because the former is very ambiguous in English. A box can be contained in a second box which can be contained in a third and we would agree that the first box is contained in the third. In some sense "$\subseteq$" is a better fit to the English "is contained in". "$\in$" could be thought of more as "belongs directly to". – user21820 Jan 31 '15 at 12:59

3 Answers3

6

Not necessarily. A set with this property is called transitive.

Joffysloffy
  • 1,344
  • 11
  • 17
6

Not always.

Suppose $A=\{a\}$, so that $a\in A$. Suppose $B = \{A\} = \{\{a\}\}$, so that $A \in B$.

However, $a \notin B$.

amWhy
  • 210,739
3

By literally following what you wrote in the title, then the answer is no.

You are saying that $a$ is a member of $A$ and $A$ is a member of $B$. The example given by @amWhy demonstrates that $B$ may not have $a$ as a member.

If you instead write $a \in A$ and $A \subset B$, then indeed $a \in B$. The first case led to $B$ having the set $A$ as a member and the second case has $A$ as a subset of $B$.

Joel
  • 16,574