If you could evaluate my proof on scale of 1-5 in clarity, conciseness, and validity, I would appreciate your 2 cent!!!

Sketch of Proof in words:

Consider for some $n\in\mathbb{N}$, $\bigcup^{n}_{i=1} P_i$ where $P_i$ is a bounded set in arbitrary metric space $(X,d)$.
By boundedness, for each $i$, $\exists x^*\in X,\exists M_i\in\mathbb{R}$, $d(x^*,p)\leq M_i$ $\forall p\in P_i$.
Let $M^*= \max\{M_i:i=1,...,n\}$.
If $q$ is in the union, then for some $i$, $q\in P_i$.
$d(q,x^*)\leq M_i\leq M^*$.
The union of any finite collection of bounded sets is bounded. QED

  • 2
    $\begingroup$ It looks clear to me; at least 4. You might include the tag proof-verification in posts like this $\endgroup$ – amrsa Nov 5 '16 at 21:37
  • $\begingroup$ @amrsa thanks. I edited the tag. $\endgroup$ – Frank Swanton Nov 5 '16 at 21:39
  • $\begingroup$ Frank: Please ignore the nonsense you see posted as an answer, upon which I commented. $\endgroup$ – Namaste Nov 5 '16 at 23:08

Assume it is unbounded. It has an unbounded sequence. It has a monotone unbounded subsequence (that is, it goes to infinity). This sequence has infinitely many terms from one of your bounded sets. This set is unbounded.

  • $\begingroup$ Say what? Please, at the very least, try to justify your answer (mathematically), or explain it? $\endgroup$ – Namaste Nov 5 '16 at 23:08

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