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Let's suppose I have a subset $A \subset \mathbb{R}$, then $A$ is bounded as $\sup A$ and $\inf A$ exist in $\mathbb{R}$.

But suppose I have $A \subset \mathbb{R}^k$, how does one define boundedness for this subset?

I've been told that $\sup A$ and $\inf A$ have no meaning in $\mathbb{R}^k$, but I can't see why this is the case, as we can impose a dictionary order on $\mathbb{R}^k$. If it is the case that $\sup A$ and $\inf A$ have no meaning in $\mathbb{R}^k$ how does one then define boundlessness for this subset?

More generally if we are in an arbitrary metric space $(X, d)$, how does one define the concept of boundedness for a subset $A \subset X$ (I know that in arbitrary topological spaces, boundedness has no meaning)

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    You have a metric. You can then have the distance from any point of $A$ to $0$, and you can take the supremum of these. – 2017-01-23
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    @ClementC In that case if A is an open ball in a metric space $X$, we will always have $\inf A = 0$ since $d(x, y) \geq 0 \ \ \forall x, y \in X$, so the condition for boundedness would simply be that $\sup A = r$ for $r \in \mathbb{R} - \infty$ – 2017-01-23
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    First of all, no: the quantity I am talking about could be strictly greater than say $27$, if all the points in $A$ are at distance at least $27$ from the origin. And second, I did not say this was a way to define either $\sup A$ or $\inf A$ (the choice of the point $0$ in my comment is natural, but arbitrary, and the definitions would depend on that point). The above is to define *boundedness*, which does not depend on the choice of $0$ as "reference point." – 2017-01-23

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Rather than thinking in terms of upper and lower bounds, think of a bounded subset $A$ of $\Bbb R$ as a set that can be put in an interval (a bounded interval that is). Now think of bounded subsets of $\Bbb R^n$ as those sets which can be put in a ball (do the same for metric spaces).

Definition: Let $(X,d)$ be a metric space and $A \subset X$. $A$ is said to be bounded if $\exists M >0$ and $x_0 \in A$ such that $A \subset B(x_0,M)$.

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A set $S \subset \mathbb{R}^k$ is bounded if $\exists$ a point s $\in S$ and some value $r > 0$ such that the open ball $B(s,r) \supset S$.

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You can define using the concept of $\sup$ too.

Definition: Let $(X,d)$ be a metric space and let $A\subset X$. We say that $A$ is bounded if $\sup\{d(x,y):x\in A, y\in A\}<\infty$. In this case, the number $\sup\{d(x,y):x\in A, y\in A\}$ is called the diameter of the set $A$.