Proof ∀a,b ∈ ℝ [ a = b ⇔ ( ∀ ϵ ∈ ℝ [ ϵ > 0 ⇒ |a − b| < ϵ ] ) ]

Prove ∀a,b ∈ ℝ [ a = b ⇔ ( ∀ ϵ ∈ ℝ [ ϵ > 0 ⇒ |a − b| < ϵ ] ) ].

We are asked to prove the equivalence:

$$ \forall a,b \in \mathbb{R},\quad a = b \iff \left( \forall \epsilon \in \mathbb{R},\ \epsilon > 0 \Rightarrow |a - b| < \epsilon \right) $$

We will prove this biconditional statement by proving both directions:

---

(⇒) Forward direction:

Assume $a = b$. We want to show:

$$ \forall \epsilon > 0,\ |a - b| < \epsilon $$

Since $a = b$, we have $|a - b| = |0| = 0$. For any $\epsilon > 0$, clearly $0 < \epsilon$. So, $|a - b| < \epsilon$.

✅ Thus, the implication holds in the forward direction.

---

(⇐) Reverse direction:

Assume:

$$ \forall \epsilon > 0,\ |a - b| < \epsilon $$

We want to show $a = b$. We’ll proceed by contradiction:

Suppose $a \ne b$. Then $|a - b| = \delta$ for some $\delta > 0$. Now, choose $\epsilon = \delta/2 > 0$. Then:

$$ |a - b| = \delta > \delta/2 = \epsilon $$

This contradicts the assumption that $|a - b| < \epsilon$ for all $\epsilon > 0$.

❌ So, the assumption $a \ne b$ must be false. Hence:

✅ $a = b$

---

✅ Conclusion:

$$ \boxed{ \forall a,b \in \mathbb{R},\quad a = b \iff \forall \epsilon > 0,\ |a - b| < \epsilon } $$