Bounded sets equivalent definition 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:
($\implies$)


*

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

*$\exists q\in X:\forall p\in E:d(p,q)<M_{0}$

*$\forall p\in E:d(p,q_{0})<M_{0}$

*$E\subset N_{M_{0}}(q_{0})$ [a neighborhood around $q_{0}$ of radius $M_{0}$]

*$\forall p,q\in E:d(p,q)<2M_{0}$ [Thanks to A.P. and Hagen for this]

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


($\Longleftarrow$)


*

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

*$\forall p,q\in E:d(p,q)<M_{0}$

*$\forall p\in E:d(p,q_{0})<M_{0}$

*$\exists q\in E:\forall p\in E:d(p,q)<M_{0}$

*$\exists q\in X:\forall p\in E:d(p,q)<M_{0}$

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

 A: 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.
