Bounded sets equivalent definition

Question

Let $X$ be a metric space, and $E\subset X$. I have two definitions of a bounded set, I want to prove they are equivalent.

Definition 1: $\exists M:\exists q\in X:\forall p\in E:d(p,q)<M$

Definition 2: $\exists M:\forall p,q\in E:d(p,q)<M$

Proof:

($\implies$)

1. $\exists M:\exists q\in X:\forall p\in E:d(p,q)<M$
2. $\exists q\in X:\forall p\in E:d(p,q)<M_{0}$
3. $\forall p\in E:d(p,q_{0})<M_{0}$
4. $E\subset N_{M_{0}}(q_{0})$ [a neighborhood around $q_{0}$ of radius $M_{0}$]
5. $\forall p,q\in E:d(p,q)<2M_{0}$ [Thanks to A.P. and Hagen for this]
6. $\exists M:\forall p,q\in E:d(p,q)<M$

($\Longleftarrow$)

1. $\exists M:\forall p,q\in E:d(p,q)<M$
2. $\forall p,q\in E:d(p,q)<M_{0}$
3. $\forall p\in E:d(p,q_{0})<M_{0}$
4. $\exists q\in E:\forall p\in E:d(p,q)<M_{0}$
5. $\exists q\in X:\forall p\in E:d(p,q)<M_{0}$
6. $\exists M:\exists q\in X:\forall p\in E:d(p,q)<M$

Answer 1

As Hagen von Eitzen said in his comment, the two definitions are equivalent only as long as $X \neq \varnothing$. Other than that, the $(\Leftarrow)$ part is correct, but the $(\Rightarrow)$ part is not. Namely, when going from 4 to 5 you should use $d(p,q) \leq d(p,q_0) + d(q,q_0) < 2M_0$ instead of $d(p,q) < M_0$.

---

From your question it isn't clear if you actually need to use logic formalism. If you don't, then you could write the proof in a more legible way, as I do below.

Here we will assume $E \neq \varnothing$ for simplicity: $\varnothing$ is always bounded by definition 2 and is bounded by definition 1 as long as $X \neq \varnothing$.

$\mathbf{1 \Rightarrow 2}:$ Let $M,q$ be as in definition 1. Then for every $p_1,p_2 \in E$ we have $$ d(p_1,p_2) \leq d(p_1,q) + d(q,p_2) < 2M $$ by the triangle inequality. Thus $E$ satisfies definition 2 with $M' = 2M$.

$\mathbf{2 \Rightarrow 1}:$ Let $M$ be as in definition 2 and fix a $q \in E$. Then by definition 2 $$ d(p,q) < M $$ for every $p \in E$. Thus $E$ satisfies definition 1, too.