- #1

playboy

Hi.

Can somebody please check my work, its this dumb proof in the text book which is the most obvious thing.

Let S and T be nonempty bounded subsets of R with [itex]S \subseteq T[/itex].

Prove that [itex]inf T \leq inf S \leq sup S \leq sup T[/itex].

I first broke it up into parts and tried to prove each part.

1. [itex]sup S \leq sup T[/itex]

2. [itex]inf S \leq sup S[/itex]

3. [itex]inf T \leq inf S [/itex]

also, define:

sup T = a

sup S = b

inf S = c

inf T = d

1. Prove that [itex]sup S \leq sup T[/itex]

By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff [itex]m \geq x [/itex]. for all [itex]x \in X[/itex]

Since S is bounded above, S has a supremum and say, sup S = b. But, [itex]S \subseteq T[/itex], and thus, [itex]b \in T[/itex].

But, by definition, a = supT iff a is (greater than or equal too) all other elements in T

Thus, [itex]a \geq b[/itex] and thus, [itex]sup T \geq sup S[/itex]

2. Prove that [itex]inf S \leq sup S[/itex]

By Definition, inf S;

c = inf S iff c is (less than or equal too) all elements in S.

and, sup S is

b = supS iff b is (greater than or euqal too) all elements in S.

Thus, since [itex]c \leq s [/itex] and [itex]b \geq s [/itex], where s is (all elements in S), [itex]c \leq b [/itex] as required.

3. This would be the same proof as 1, but using the Definiton of infimum instead.

So can somebody please check this proof and if I am doing it properly?

Can somebody please check my work, its this dumb proof in the text book which is the most obvious thing.

Let S and T be nonempty bounded subsets of R with [itex]S \subseteq T[/itex].

Prove that [itex]inf T \leq inf S \leq sup S \leq sup T[/itex].

I first broke it up into parts and tried to prove each part.

1. [itex]sup S \leq sup T[/itex]

2. [itex]inf S \leq sup S[/itex]

3. [itex]inf T \leq inf S [/itex]

also, define:

sup T = a

sup S = b

inf S = c

inf T = d

1. Prove that [itex]sup S \leq sup T[/itex]

By Definition, let X be a nonempty subset of R and let X be bounded above. Thus, m = sup X iff [itex]m \geq x [/itex]. for all [itex]x \in X[/itex]

Since S is bounded above, S has a supremum and say, sup S = b. But, [itex]S \subseteq T[/itex], and thus, [itex]b \in T[/itex].

But, by definition, a = supT iff a is (greater than or equal too) all other elements in T

Thus, [itex]a \geq b[/itex] and thus, [itex]sup T \geq sup S[/itex]

2. Prove that [itex]inf S \leq sup S[/itex]

By Definition, inf S;

c = inf S iff c is (less than or equal too) all elements in S.

and, sup S is

b = supS iff b is (greater than or euqal too) all elements in S.

Thus, since [itex]c \leq s [/itex] and [itex]b \geq s [/itex], where s is (all elements in S), [itex]c \leq b [/itex] as required.

3. This would be the same proof as 1, but using the Definiton of infimum instead.

So can somebody please check this proof and if I am doing it properly?

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