EXAMPLES
-
Prove that the set R+ of positive real numbers is not bounded above.
Sol. The set R+ of positive real numbers is not bounded above. For, suppose that R+ is bounded above and u is an upper bound.
Since 1 ∈ R+ and u is an upper bound for R+, therefore.
1 ≤ u, it means u > 0.
u + 1 > 0 and consequently u + 1 ∈ R+.
Thus there exists u + 1 ∈ R+ such that u + 1 > an upper bound(u) of R+. But this is a contradiction because if u is an upper bound for R+, then we must have x ≤ u for all x in R+.
Hence u is not upper bound of R+ and so R+ is not bounded above.
2. Prove that the set R- of negative real numbers is not bounded below.
Sol. The set R- of negative real numbers is not bounded above. For, suppose that R+ is bounded above and u is an upper bound.
Since -1 ∈ R- and v is a lower bound for R-, therefore.
-1 ≥ v, it means v < 0.
v – 1 < 0 and consequently
v – 1 ∈ R-.
Thus there exists v – 1 ∈ R- such that v – 1 < a lower bound(v) of R-. But this is a contradiction because if v is a lower bound for R-, then we must have x ≥ v for all x in R-.
Hence v is not lower bound of R- and so R- is not bounded below.
3. Find the supremum and infimum of the singleton.
{2} ⊂ R.
Sol. For {2} ⊂ R, we find that the set of upper bounds of {2} is given by {x ∈ R : x ≥ 2 } and 2 being the least of these upper bound bounds, we have, sup {2} = 2.
Again the set of lower bounds of { 2} is given by
{ x ∈ R : x ≤ 2 }
And 2 being the greatest of these lower bounds, we have
Inf{2} = 2
Hence sup{2} = 2 = inf{2}.
5. Find the supremum and infimum
{(1/n) : n ∈ N}
Sol. Let S = {(1/n) : n ∈ N}
S = { 1, ½, 1/3, …………… }
Clearly, S is bounded because 0 ≤ x ≤ 1 [ x ∈ S. inf S = 0 ∉ S and sup S =1 ∈ S. Here the greatest member of S is 1 whereas the smallest member of S does not exists.
​
​