Definition of bounded in a metric space - confirmation

Question

Is the following definition of a bounded metric space correct?

$(M,d)$ is bounded if $\exists a \in M, r > 0$ such that $M = B(a,r)$.

Looking around on the internet I instead see $M \subset B(a,r)$.

Answer 1

The two definitions are equivalent. In particular, since a ball is defined as $$B(a,r)=\{x\in M:d(x,a)<r\}$$ it is trivial that we have $B(a,r)\subseteq M$ for any $a$ and $r$. Knowing this, the statement that $M\subseteq B(a,r)$ implies that $B(a,r)=M$ since $\subseteq$ is an antisymmetric relation.

However, one might note that if you want to define a bounded subset $S\subseteq M$, then you would write $S\subseteq B(a,r)$ rather than $S=B(a,r)$, since the ball would be taking place in $M$ rather than intrinsically $S$.

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The definition $M\subseteq B(a,r)$ is a good definition for a metric space or subset thereof being bounded. This coincides with the intuition people want to capture by boundedness, though it is equivalent to other definitions. Moreover, in the definition $M=B(a,r)$, one could easily forget that the ball on the right hand side of the equation must be taken with respect to $M$ and not to some larger space, where writing $M\subseteq B(a,r)$ does not allow one to make such a mistake.