A simpler (o rather direct) proof for $P(\{a\})=\big\{ \emptyset, \{a\}\big\}$.

Question

Here's my proof but I'm looking for a simpler or at least direct proof for the indirect part of this proof.

Proof. Given two sets, say $a$ and $x$, it all boils down to stablish.\begin{align*}x\subseteq \{a\}\iff x=\emptyset\vee x=\{a\}\end{align*} ($\impliedby$)Trivial.

It's trivial once we know $\forall S: S\subseteq S\wedge \emptyset \subseteq S$

($\implies$)Let us assume now, $$x\subseteq \{a\}\wedge x\ne \emptyset\tag{*}$$If there is $t\in\{a\}: t\notin x$ by the uniqueness of the empty set we know $x=\emptyset$, a contradiction to $(*)$ So it must be that $\forall t:\big(t\in \{a\}\implies t\in x\big)$. Furthermore, we know $\forall t:\big(t\in \{a\}\iff t\in x\big)$ also because of $(*)$Then,\begin{align*}\Big((*)\implies x=\{a\}\Big)&\iff \Big(x\subseteq \{a\}\implies \big(x\notin \emptyset\implies x=\{a\}\big)\Big)\\ &\iff \Big(x\subseteq \{a\}\implies x=\emptyset \vee x=\{a\}\Big)\end{align*} as we intended to prove.

Answer 1

Similar to the proof by Hermis14 but more direct, avoiding contraposition. As already observed, it suffices to prove that $\forall x\,(x\subseteq\{a\} \implies(x=\varnothing \lor x=\{a\}))$. So consider any $x\subseteq \{a\}$. Thus all members of $x$ (if any) are equal to $a$.

Case 1: $a\in x$. Then $a$ is the one and only member of $x$, so $x=\{a\}$.

Case 2: $a\notin x$. Then $x$ has no members at all, so $x=\varnothing$.

In both cases, $x=\varnothing \lor x=\{a\}$ as required.

Answer 2

$ \newcommand{\ps}{\mathcal{P}} \newcommand{\eqv}{\Leftrightarrow} \newcommand{\emt}{\varnothing} $ By the definition of the power set, $$ \forall x: (x \in \ps(\{a\}) \eqv x \subseteq \{a\}) $$ Let $P = \{\emt, \{a\}\}$. Then, we have to show that $$ \forall x: (x \subseteq \{a\} \eqv x \in P) $$ to show $P = \ps(\{a\})$. Since you showed the $\Leftarrow$ part, I will show the other direction proof by contraposition: $$ \forall x: (x \notin P \Rightarrow x \not\subseteq \{a\}) $$ Suppose $x \notin \{\emt, \{a\}\}$. Then, $x \neq \emt \land x \neq \{a\}$. There must be $b$ such that $b \in x \land b \neq a$. Hence, $x$ is not a subset of $\{a\}$.