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I don't know I'm misinterpreting, but I'm a little confused about the definition of bounded. As far as I was aware, a bounded set of real numbers is one that is bounded both below and above. Therefore could an unbounded set of real numbers be one that is either exclusively bounded below or above or one that is both unbounded below and above?

Reason for the question is that I was working on a proof of the statement that a non-sequentially compact set of real numbers either has (i) an unbounded sequence in S or (ii) there is a sequence in S that converges to a point $x_0 \notin S$.

So $S$ could be a bounded set or an unbounded set.

In the case that it is unbounded, I think it seems pretty clear that we can construct an unbounded sequence, but if the sequence converges to a lower or upper bound (assuming the set only has one of them) then we could have that statement (ii) is true.

I thought this was pretty solid until I found a solution online that seemed to indicate that a set of the form $(a, \infty)$ or $(-\infty,a)$ is considered bounded.

So for a set of real numbers, is the only option for an unbounded set $(-\infty, \infty)$? Does this change depending on who you talk to or is the accepted definition?

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    $\begingroup$ Both $(-\infty,a)$ and $(a,\infty)$ are unbounded. $\endgroup$ – TorsionSquid Jul 9 '15 at 2:36
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    $\begingroup$ Another example where a modifier widens rather than restricts: "bounded above" and "bounded below" are less restrictive than "bounded". $\endgroup$ – Robert Israel Jul 9 '15 at 3:49
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Bounded in a metric space means that the set can be contained in a ball of fixed radius. Often the qualifiers bounded above or below are used when the elements are allowed to be arbitrarily large negative or positive reals, respectively.

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If we let $S=(-\infty,a)\cup(a,\infty)$ for some $a\in\Bbb R,$ then $S$ is unbounded.

However, a bounded set can fail to be sequentially compact. For example, given $a,b\in\Bbb R$ with $a<b,$ we have that $(a,b)$ is bounded, but fails to be sequentially compact, because it satisfies condition (ii).

Further, note that if a set $S\subseteq\Bbb R$ has a lower bound or an upper bound but not both, then $S$ necessarily satisfies condition (i), but need not satisfy condition (ii). For example, consider $[0,\infty).$

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A bounded set of Real numbers is one that can be contained in an interval $(a,b)$, for $-\infty <a<b < \infty $. Equivalently, in $\mathbb R^n$ , a bounded set is one that can be contained in a ball of finite radius.

Equivalently, a subset $S$ of $\mathbb R^n$ is bounded iff, $\forall x,y \in S, d(x,y)<\infty $, or $Sup$ {$d(x,y) : x,y \in S$} $< \infty$, or $d: S \times S \rightarrow \mathbb R$ is a bounded function, where $d(x,y)$ is the standard Euclidean metric.

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