Is a Bounded Set Always Finite or Can It Go to Infinity?

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A bounded set is defined as a set that is contained within some ball, and it is not synonymous with a set that goes to infinity. A set can be bounded even if it contains an infinite number of members, such as the set {1/1, 1/2, 1/3, ...}, which is bounded above and below. The concept of boundedness is independent of whether a set is open or closed, and a set can have multiple bounds that are not members of the set itself. To be considered bounded, a set must have both a lower and an upper bound.

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Is a bounded set synonymous to a set that goes to infinity? I feel like unless a set is
(-infinity, n) or [n, infinity) it is not going to be unbounded.

The other thing that I was wondering is can a set be neither open nor closed AND unbounded? Doesn't the definition of open/closed imply that there is a boundary?

Thanks!
 
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A set being bounded just means it is contained in some ball. Boundedness isn't really related to being open or closed, and also has nothing to do with boundary.
 
Is a bounded set synonymous to a set that goes to infinity?

No and furthermore the bound does not even have to be a member of the set.

A set can have lots of bounds, even an infinite number of them.

A bound can be 'above' or 'below'.

An upper bound is simply a real number that is either greater than or equal to every member of the set or less than/equalto every member of the set.

so 7,8,9,10 etc all form upper bounds to the set {2,3,4,5,6} ; none are memebrs of the set.
but wait 6 also forms an upper bound as it satisfies the equal to and is a member.

Similarly 1,0,-1,-2 all form lower bounds that are not members and 2 forms a lower bound that is

Sets which 'go to infinity' are unbounded. However a set can contain an infinite number of members and still be bounded, above and/or below.

for example the set {1/1, 1/2, 1/3, ...} is bounded above by 2 ,1 etc and bounded below by 0, -1 etc, but contains an infinite number of members.

The set \{ - \infty ,... - 2, - 1,0,1,\frac{1}{2},\frac{1}{3},...\}

is not bounded below but is bounded above.

If a set has both a lower and upper bound so that the modulus of any member, x, is less than or equal to some real number K then the set is bounded. (No upper or lower)

If for any x \in S there exists a K such that

\left| x \right| \le K

The set S is bounded.

Hope this helps.
 
"bounded" does not mean "has a boundary"

Isn't English confusing?
 
So I guess what I'm really wondering is can a set be unbounded if it doesn't go to infinity? That's what I can't seem to wrap my brain around, because I feel that if a set doesn't go to infinity, there will always be a real number K larger than the members of the set.
 
Any finite set S of real numbers is bounded.
Infinite sets can be bounded or unbounded.

Remember that to be bounded a set must have both a lower bound and an upper bound.
Sets with only a lower or upper bound are unbounded.

A further question for you to ponder:

Can a set of complex numbers be bounded?
 
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