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Given a sequence $\left(x_n\right) \in \mathbb{R}$ which converges to $l \in \mathbb{R} $ where $$ \alpha<x_n \text{ for all } n \in \mathbb{N}.$$ show that $\alpha \leq l$.

I tried to prove this. I got an idea but I cannot express myself mathematically which is 100% correct. Here is my proof, Suggest a Correct Mathematical Proof for pure maths students. Thank You.

Attempt

Given $\left(x_n\right)$ converges to $l$ $\implies$ for any $\epsilon >0$ there exists an $n_0\left(\epsilon\right)$ such that for all $ n\geq n_0\left(\epsilon\right)$ we have $l-\epsilon < x_n < l+ \epsilon$ because $\alpha < x_n$, this we have $\alpha< x_n<l-\epsilon$ for all $ n\geq n_0\left(\epsilon\right)$.

Hence we have $\alpha- l<\epsilon$ hence $\alpha- l$ cannot take positive values hence $\alpha- l \leq 0$ hence the conclusion.

But I am not sure if I had provided every required details, I need someone to make it Logically and Mathematically correct.

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If $\alpha >l$ take $\epsilon =\alpha -l$ to get a contradiction.

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  • $\begingroup$ Didn't get you. Can you please give a better solution $\endgroup$ – user581912 Aug 9 '18 at 0:21
  • $\begingroup$ You have already provided all the necessary details except for getting a contradiction at the end. Note that $x<x$ is not true for any real number $x$. Hence $\alpha-l<\epsilon =\alpha-l$ is a contradiction proving that $\alpha >l$ is impossible. $\endgroup$ – Kavi Rama Murthy Aug 9 '18 at 0:24
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You know that for each $\epsilon>0$, there is an $n_{\epsilon}$ such that for $n\geq n_{\epsilon}$, $\alpha<x_{n}<l+\epsilon$. Thus, $\alpha<l+\epsilon$. But this holds for all $\epsilon$, so it must be that $\alpha\leq l$.

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Suppose for a contradiction that $\alpha > l$. Then $\varepsilon = \alpha - l > 0$. Since $x_n \rightarrow l$, therefore $|x_n - l| < \varepsilon = \alpha - l$ for sufficiently large $n$. Then for sufficiently large $n$

$x_n - l \leq |x_n - l| < \alpha - l \Longrightarrow x_n < \alpha$

This contradicts the fact that $x_n > \alpha$ for all $n$.

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$\alpha \lt x_n \forall n .$ Now,$(x_n)\to l\implies \exists$ finitely many n such that $ x_n \not\in(l-\varepsilon,l+\varepsilon)\therefore\alpha\lt x_n(\forall n) \le l-\varepsilon$(for finitely many n)$ \lt l, $.That is $\alpha\lt l$. The case,$\alpha=l$ may occur when,$l \lt x_n \forall n$,i.e. $(x_n)$ is decreasing and doesn't attain its limit($x_n \ne l $,for any n).For example,$\left(\dfrac{1}{n}\right)\to 0$,but any term of the sequence never $=0$.In such case,we may have $\alpha=l$

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