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## Homework Statement

The question : http://gyazo.com/e2dfc19ce0bc2ff857019edfc3f42edb

## Homework Equations

Supremum and infimum axioms.

## The Attempt at a Solution

Suppose ##A = \{p,q \space | \space p,q \in ℝ, \space p ≠ q\}## is a bounded subset of ##ℝ##.

**(a)**Since ##A## is bounded, we know it is bounded above by p, q or ∞ and we also know it is bounded below by p, q or -∞

W.L.O.G, lets assume that ##p > q##. So ##p## is an upper bound for ##A##, but ##A## is also bounded above by ##∞##. Since ##p < ∞##, we may write ##sup(A) = p < ∞##.

Similarly, ##q## is a lower bound for ##A##, but ##A## is also bounded below by ##-∞##. Since ##p > -∞##, we may write ##inf(A) = q > -∞##.

Putting these together we observe ##-∞ < inf(A) < sup(A) < ∞##.

**(b)**Suppose ##B## is a nonempty subset of ##A##.

Since ##inf(A)## is a lower bound of ##A##, we have ##inf(A) ≤ a, \space \forall a \in A##.

In particular, ##inf(A) ≤ b, \space \forall b \in B## since ##B \subseteq A##. Thus ##inf(A)## is a lower bound of ##B##, so ##inf(A) ≤ inf(B)##.

Picking any ##b \in B## shows that ##inf(B) ≤ b ≤ sup(B)##, which gives the second inequality.

Finally, since ##sup(A)## is an upper bound of ##A##, we know ##\space a ≤ sup(A), \space \forall a \in A##.

Hence ##b ≤ sup(A) \forall b \in B \Rightarrow sup(A)## is an upper bound of ##B##. Thus ##sup(B) ≤ sup(A)## since ##sup(B)## is the least upper bound of ##B##.

Therefore putting these all together, we have our desired inequality.

**(c)**Suppose ##B = \{ b \in ℝ \space | \space b ≥ a, \space \forall a \in A \}##

Since ##A ≠ ∅## and is bounded, we know ##sup(A)## exists so ##B ≠ ∅##. ( Don't think this is right ).

Suppose ##a \in A##. Since ##a ≤ b, \space \forall b \in B##, we know ##b## is a lower bound of ##B##, so ##B## is bounded below.

I'm having a bit of trouble proving ##inf(B) = sup(A)## at this point.