# Can a set be infinite and bounded?

I don't understand a statement in my math book course, I was restudying the compact sets part of the chapter when at a certain moment there is a corollary saying :

'every infinite and bounded part of $\mathbb{R^n}$ admit at least one accumulation point'

because for me a set is either bounded so finite or infinite so unbounded.

I don't really understand because I can accept the fact that without a metric, bounds make no sense in topology but here $\mathbb{R^n}$ is clearly known as a metric space.

thank you for your help

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The set $\{1,1/2,1/3,1/4,\dots\}$ is infinite and bounded. Bounded means there is a "ball" that contains the set. – André Nicolas Aug 7 '14 at 5:30
Do you know what the definition of bounded is? – Andrés Caicedo Aug 7 '14 at 5:30
An accumulation point of @AndréNicolas set is 0 if choose to traverse it from left to right – frogeyedpeas Aug 7 '14 at 5:31
Any interval is infinite but bounded. – Lucyfer Zedd Aug 7 '14 at 5:40
I thought an infinite set was about how 'big' it was in a sense like : $]a;b[$ was finite and $]a;\infty[$ infinite but i just understood what you meant – Brocolus Aug 7 '14 at 5:43

When we say that a set is finite or infinite, we are referring to the number of elements in the set, not to the "extent" (putting it roughly) of those elements. You can think of it in the following way. Any set, all of whose elements lie between (for example) $0$ and $1$, is bounded, because no part of the set can possibly "go to infinity". But clearly it is possible to have an infinite number of elements in such a set. For example, all reals between $0$ and $1$, or all rationals between $0$ and $1$, or simply all the numbers $\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\,\}\,$.

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oh right that's exactly where my problem was, i thought the fact the set is countable or not countable was not linked to the finite or infinite. Thank you very much ! – Brocolus Aug 7 '14 at 5:39
Countability is a different concept altogether. An infinite bounded set can be countable (e.g. all rationals between 0 and 1) or uncountable (e.g. all reals between 0 and 1). – BenM Aug 8 '14 at 0:45
Oh Yea true i'm saying nonsense – Brocolus Aug 8 '14 at 8:13

The set of all numbers between $0$ and $1$ is infinite and bounded. The fact that every member of that set is less than $1$ and greater than $0$ entails that it is bounded.

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Leibniz speculated about intervals (in time and/or space) that are simultaneously infinite and also terminating (one way of interpreting "bounded*). The presence of such sets in his calculus is what led him to consider infinite (and infinitesimal) numbers as useful fictions, to the chagrin of many of his disciples who held a more realist view of infinitesimals.

In a modern hyperreal framework, you can indeed have an infinite interval $[0,H]$ which is nontheless bounded (by an infinite hyperreal).

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infinite refers to the cardinal of the set.

bounded refers to the values.

$(1, \frac{1}{2},\frac{1}{3},\frac{1}{4} ...)$ is a set that defines an infinite number of values, all of which are included in the interval $(0, 1]$

It is infinite and bounded.

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