Darío

Yo have reason, in the first i think as follow (a) Proof Assume that [latex]a[/latex] is the least element of [latex]B[/latex] in [latex]R^{-1}[/latex], then [latex]\forall x\in{B},a\leq{x}[/latex] Now, for [latex]R[/latex] is [latex]\forall x\in{B},x\leq{a}[/latex] Hence [latex]a[/latex] is the greatest element of [latex]B[/latex] in [latex]R[/latex]. The reciprocal proof is similary... (b) Proof For Supremum... Assume that [latex]a[/latex] is the supremum of [latex]B[/latex] in [latex]R^{-1}[/latex], then [latex]\forall x\in{B},x\leq{a}[/latex] For R is [latex]\forall x\in{B},a\leq{x}[/latex] This prove that [latex]a[/latex] is lower bound of [latex]B[/latex] in [latex]R[/latex]. Now only remains to prove that [latex]a[/latex] is the mininum lower bound... How i prove this? Good day

Hi, i should be this proof Let [latex]R[/latex] be an ordering of [latex]A[/latex]. Prove that [latex]R^{-1}[/latex] is also an ordering of [latex]A[/latex], and for [latex]B\subset{A}[/latex], (A) [latex]a[/latex] is the least element of [latex]B[/latex] in [latex]R^{-1}[/latex] if and only if [latex]a[/latex] is the greatest element of [latex]B[/latex] in [latex]R[/latex]. (B) Similarly for (minimal and maximal) and (supremum and infimum. Good look!

Is required to prove the derivative of [latex]e^x[/latex] by definition, so [latex]\displaystyle\frac{d}{dx}(e^x)=\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x +x}-e^x}{\Delta x}}=e^x\lim_{\Delta x \to 0}{\displaystyle\frac{e^{\Delta x}-1}{\Delta x}}[/latex] I can also express the exponential function as [latex]e^{\Delta x}=\sum_{n=0}^{\infty}{\displaystyle\frac{{\Delta x}^n}{n!}}[/latex] [latex]e^{\Delta x}-1=\sum_{n=1}^{\infty}{\displaystyle\frac{{(\Delta x)}^n}{n!}}=\sum_{n=0}^{\infty}{\displaystyle\frac{{(\Delta x)}^{n+1}}{(n+1)!}}[/latex] How i can continue?

This is equivalent to prove that the limit is unique, something trivial... Thanks!

I know that the sequence can write as [latex]a_n=\begin{cases} 1\ if\ n\ is\ even \\ 3\ if\ n\ is\ odd \end{cases}[/latex] So the sequence is osillating, therefore not exist the limit. But i need prove this formally, then i need the definition. how i can the rigorous definition?

Hi forum! I have that the sequence [latex]a_n=\{2-(-1)^n\}[/latex] not converges. I must show this with the rigorous definition. I think use [latex]\exists{\epsilon>0}\forall{N\in\mathbb{N}}\exists{n\ge N}:|a_n-\ell|\ge\epsilon[/latex] How i can continue?