# What does it mean for a set to be contained in another?

If $$A$$ & $$B$$ are two sets, what does '$$A$$ is contained in $$B$$' mean ? Does it mean that $$A$$ is a subset of $$B$$ or $$A$$ is a proper subset of $$B$$?

• There is no universal convention, check the one used by the particular author. – Yves Daoust Jan 12 at 15:28

It's a bit ambiguous, but hopefully it's only used when there is little chance of confusion. If there is a chance, it is better to use notation. Either $$A\in B$$ for $$A$$ being an element of $$B$$, or $$A\subseteq B$$ for $$A$$ being a subset of $$B$$.

If it's the latter, generally it is not necessarily a proper subset. Authors however can do whatever they want, as long as they define it. You have to watch out for nonstandard language and notation.

• @Thomas Started writing it on my phone, had to do something else, must've accidentally posted it without finishing. Thanks. – Matt Samuel Jan 12 at 15:27
• I would go as far as saying that writing "$A$ is contained in $B$" and meaning $A\in B$ amounts to gaslighting... – JuliusL33t Jan 12 at 15:56
• @Julius Sets can contain elements. And in set theory everything is a set. So in situations like that it can legitimately be ambiguous. – Matt Samuel Jan 12 at 16:19
• I'm not commenting on the mathematical content, but on pedagogy and custom. Some people define "cat" to mean "dog" and "up" to mean "down". Saying "$A$ is contained in $B$" and meaning "$A$ is an element of $B$" seems so cruel to me, especially when this is very standard terminology. But hey, whatever floats your boat. – JuliusL33t Jan 12 at 16:36
• I DO NOT disagree with you answer, so don't take offense. – JuliusL33t Jan 12 at 16:39

Because we do know nothing about the two sets let me give you two examples.

1) We know that $$N \subset R$$. This means that $$N$$ is contained in $$R$$.

2) Let us define $$A = \{x,y\}$$ and $$B = \{x,y\}$$ then A is contained in B and B is contained in A.

The only thing that "$$A$$ is contained in $$B$$" means is that the implication
$$x \in A \implies x \in B$$