2

According to this post, we need the axiom of separation to show that $\{x,x\} = \{x\}$.

I struggle to see why this is so. It seems to me the axiom of pairing allows us to construct a set $X$ such that

$$y\in X \iff (y = x \lor y = x)$$

yet the latter condition is equivalent in first-order logic to $y=x$, hence

$$y\in X \iff y = x.$$

Sam
  • 5,208
  • 2
    Things don't exist in a vacuum. You need to tell us what the other axioms first. Pairing isn't always an axiom, since pairing itself is a consequence of Replacement and Infinity, or Replacement and Power Set and Empty Set, or Replacement and Power Set and Infinity. And so on. – Asaf Karagila Nov 26 '24 at 10:44

1 Answers1

2

Pairs axiom: $∀x∀y∃z∀w[w ∈ z ↔ (w = x \lor w = y )]$ is enough.

$\{ x \}$ is simply $\{ x,x \}$.

As per comment below, we can use a weaker form of the Pairs axiom: $∀x∀y∃z[x∈z ∧ y∈z]$, in which case we know that the set $z$ has $x$ and $y$ as elements, but we cannot rule out the fact that there are others.

Thus, we need Separation with formula $(t=x ∨ t=y)$ to carve out the pair from $z$.

Mauro ALLEGRANZA
  • 99,247
  • 2
    Doesn't the need for separation depend on whether the axiom of pairing is formulated in the elaborate form $$\forall x\forall y\exists z\forall t[t\in z\leftrightarrow(t=x\lor t=y)]$$ or in the simple form $$\forall x\forall y\exists z[x\in z\land y\in z]?$$ – user14111 Nov 26 '24 at 08:19