I am tasked with prove the following elementary result. I am concerned about being rigours enough in my proof:
$$a\in S \iff \{a\} \subseteq S $$
My Attempt:
Suppose $\{a\} \subseteq S $ . Then every element of $\{a\}$ would be a element of $S$ but $ a$ is the only element of $\{a\}$ thus $a \in S$
While I have no qualms about this being true, I am unsure on how to preceed with in a rigorous and verbose manner
I would think the constructing the set $\{a\}$ and noting that $a \in S$( by Hpothesis) so every element of $\{a\}$ would be an element of $S$ and thus $\{a\} \subseteq S $ would be the way to go.
$\endgroup$ 34 Answers
$\begingroup$What you said is basically right, although you want to make sure to prove both directions of the implication. So,
$\endgroup$ $\begingroup$By the definition of the subset relation, $\{a\}\subseteq S$ if and only if every element of $\{a\}$ is an element of $S$. By the definition of the singleton set, the only element of $\{a\}$ is $a$. Therefore, $\{a\}\subseteq S$ if and only if $a\in S$.
$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & } \newcommand{\endcalc}{\end{align}} $I would prove this rigorously as follows: $$\calc \{a\} \subseteq S \calcop{\equiv}{definition of $\;\subseteq\;$} \langle \forall x : x \in \{a\} : x \in S \rangle \calcop{\equiv}{definition of $\;\{\ldots\}\;$} \langle \forall x : x = a : x \in S \rangle \calcop{\equiv}{logic: one-point rule} a \in S \endcalc$$
$\endgroup$ 2 $\begingroup$Suppose that $\{a\}\subset S$. We have that $a\in\{a\}$. Since $\{a\}\subset S$, it follows that $a\in S$.
Suppose that $a\in S$. Take any $x\in\{a\}$. It follows that $x=a$. Since $a\in S$, it follows that $x\in S$. We have shown that $\forall x\ (x\in\{a\}\implies x\in S)$, so $\{a\}\subset S$.
If this is not what you need, then perhaps you may find inspiration here:
$\endgroup$ $\begingroup$$\textbf{Proposition. } a\in S \iff \{a\} \subseteq S $
$\textit{Proof.}$
Let $S$ be a set and let $a\in S$. Consider the set $\{a\}$. Now take the element $a\in \{a\}$. Since $a\in S$ and $a$ is the only element of $\{a\}$ it follows that $\{a\} \subseteq S$. Thus, $a\in S \Rightarrow \{a\} \subseteq S$
Let $\{a\} \subseteq S$. Set-inclusion is defined such that the set $B$ is contained within the set $C$ (i.e. $B\subseteq C$) if and only if $\forall b\in B,b\in C$. Then, since $a\in\{a\}$, it follows that $a$ must be in $S$. Thus, $\{a\} \subseteq S \Rightarrow a\in S$.
Therefore $a\in S \iff \{a\} \subseteq S $.
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