Prove that x $\in y$ iff $\{x\}\subseteq y$

Question

My goal is to prove that $x\in y$ iff $\{x\}\subseteq y$.

Here's my sketch of a proof:

We have $x\in y$, and by definition $A\subseteq B := \forall a$ ($a\in A \implies a \in B$). We may infer $\forall$$z$ ($z = x$ $\implies$ $x \in y$) and $\forall$z ($z = x$ $\iff$ $z \in {x}$), then $\forall$$z$ ($z \in \{ x\}$ $\implies$ $z \in y$). And i stopped here.

I'm pretty sure that there is a simpler way of writing this proof (and by simpler, i mean doing this demonstration without invoking first order logic formalizations of set theoretic definitions)

Answer 1

Your proof looks fine to me for the forward direction! If you want a proof in words instead of logical symbols:

• First suppose $x \in y$ and we want to show $\{x\} \subseteq y$. So fix any $z \in \{x\}$; by definition this means $z = x$, therefore $z \in y$.

For the backward direction:

• Suppose $\{x\} \subseteq y$. Then since $x \in \{x\}$, by the definition of subset, $x \in y$. $\qquad \square$