Sets and subsets

Sets and subsets

Now that we know what sets are, how they are represented, and which sets are the same, we can talk about another important relationship between them. That is, one set being the subset of another set. If we have a set, say $S$, and a set, say $Q$, then $S$ is a subset of $Q$ if every member of $S$ is also a member of $Q$. So for example, if $S=\{a, b,c\}$ and $Q=\{a, b, c, d\}$, then $S$ is a subset of $Q$.

One thing that is important to note with the subset relation is that it does not exclude the possibility that two sets are the same. Indeed, there is a general fact concerning this matter. Namely, if $S$ and $Q$ are sets, and $S \subseteq Q$ and $Q \subseteq S$, then $S=Q$. In other words, if $S$ is a subset of $Q$, and $Q$ is a subset of $S$, then $S$ and $Q$ are identical.

Now if we wanted to specify explicitly that one set is a subset and not equal to another set, we can use the symbol $\subset$, which stands for ‘proper subset’. Proper subsets are just like subsets, except they come with the additional caveat that the two sets are not the same. In other words, $S \subset Q$ is just a short way to say that $S \subseteq Q$ and $S \neq Q$ (it is not the case that $S = Q$).

One thing that usually trips up people new to set theory (and sometimes, even those who aren’t) is differentiating between $\in$ and $\subseteq$, that is, differentiating between one set being a member of another set, and one set being a subset of another set. It’s important to make sure you can distinguish between the two!

Finally, we must mention the case of $\emptyset$. Though it may sound a bit strange at first, $\emptyset$ is a subset of every set, including itself! Why? Because all its members are members of every other set. Why? Because it has none! Thus, in general, for every set $S$, $\emptyset \subseteq S$. On the other hand, it is not the case that every set has $\emptyset$ as its member. Some sets may, some sets may not. So for example, if we take $S=\{\{\emptyset\}\}$, $\emptyset$ is not a member of $S$, but $\emptyset \subseteq S$. Moreover, $Q=\{\emptyset\}$ is a member of $S$, and $\emptyset \subseteq Q$ and $\emptyset \in Q$.