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.