any thoughts? this seems easy but i want a 100% proof

[Anonymous No.9177334]

Let A be a non empty subset of [math]\mathbb{R}[/math]. Suppose that there exists a sequence of real numbers [math]x_n[/math] such that:
i) every term of [math]x_n[/math] is an upper bound of A
ii) [math]x_n \rightarrow a[/math] for some [math]a \in A[/math].

prove that [math]a=max\{A \}[/math].

[Anonymous No.9177347]

if it seems easy what's your proof?

[Anonymous No.9177352]

I'm confused what exactly you're asking for, OP.

[Anonymous No.9177355]

>>9177334
Proof by contradiction:
suppose sup(A) > a
then there is exists b > a in A

define \epsilon := (b-a)/2

since x\_n -> a there exists some i so that | x\_i - a | < \epsilon. then x\_i < b which contradicts the assumption i)

[Anonymous No.9177358]

>>9177355
i can't into formating

[Anonymous No.9177359]

>>9177355
Starting from the assumption that sup(A) > a is wrong.

[Anonymous No.9177361]

>>9177347
as a non empty subset of [math]\mathbb{R}[/math] A has a supremum [math]s:=supA[\math]. So: [math]\forall \epsilon >0 , \exists a \in A: s - \epsilon < a< s[/math]. Also, [math]\forall \epsilon >0 \exists n_0 (\epsilon): \forall n>n_0 (\epsilon) : \left|x_n - a \right|<\epsilon[/math]. Rewritting the inequalities, and using the fact that [math]x_n[/math] is an upper bound of A we can show that [math]a=s=supA[/math] and because [math]a=supA \in A[/math] , [math]a=max\{ A \}[/math] . QED?

[Anonymous No.9177362]

>>9177355
Ignore me >>9177359, I must have just had a stroke.

[Anonymous No.9177364]

>>9177362
kek