Is $\subset$ a total order on the set $A = \{\{a\}\}$?

Question

Let $A = \{\{a\}\}$ be a set and consider the relation $\subset$ on A. Is $\subset$ an example of a total order on $A$?

I was thinking that since $a \not\subset a$ it wouldn't be an example of total order, but I don't know if $\emptyset$ could be used here. In which case $\emptyset \subset a$ and so it would be a total order?

Furthermore, if this was true, should the set $A$ be defined as $A = \{\emptyset, (a)\}$ instead?

Answer 1

Here is a general hint to any math problem you encounter. Always go back to first principles, by which I mean, always go back to definitions. It is important that we identify what we are trying to do and what do we need in order to do that.

Let us recall here the definition of a total order.

Let $X$ be a set. A total order on $X$ is any binary relation $\leq$ on $X$ such that

$(1) \;$ $\leq$ is reflexive, i.e. $\forall x \in X, \; x \leq x$.

$(2) \;$ $\leq$ is transitive, i.e. $\forall x, y, z \in X, \; x \leq y \wedge y \leq z \Rightarrow x \leq z$.

$(3) \;$ $\leq$ is antisymmetric, i.e. $\forall x, y \in X, \; x \leq y \wedge y \leq x \Rightarrow x = y$.

$(4) \;$ $\leq$ is strongly connected (or total), i.e. $\forall x, y \in X, \; x \leq y \vee y \leq x$.

Now, let’s consider our specific problem. Let $A = \{\{a\}\}$ (since by $(a)$ you mean the singleton that contains $a$, I will adhere to the usual notation $\{a\}$) and let’s consider the binary relation $\subset$ on $A$.

In fact, the binary relation $\subset$ on $A$ will fail to be reflexive, since $x \not\subset x$ for any set $x$. A good example would be to work with the binary relation $\subseteq$ on $A$.

The first three points establishes that $\subseteq$ is a partial order on $A$. You can easily prove this for the general case, by following the meaning of $\subseteq$. for this specific case, it is even easier to check these points, since we only have to deal with the set $\{a\}$.

Now, for the last point, which makes this partial order on $A$ a total order on $A$. Let $x, y \in A$. Note that $x = \{a\} = y$ and so in this case we have that $x \subseteq y$ and $y \subseteq x$. So, clearly we have that $x \subseteq y$ or $y \subseteq x$.

Therefore, we have that $\subseteq$ is a total order on $A$.

For the second part, concerning the $\emptyset$. I’m quite sure you don’t have to think of it at all, since it is not a element of our set $A$ and it does not figure in any of the above conditions regarding the definition of a total order on a set.

I hope this helps you with your study.