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Can anyone check my proof. Im new to writing proof and i'm not really confident about it. Any feedback is appreciated thank you.

Show that there exists two sequence strictly increasing such that $\{x_n\}_n$ is a sequence of rational numbers and $\{y_n\}_n$ is a sequence of irrational numbers that both converge to a real number $x$.


In the statement, we know that $\lim_{n \to \infty} \{x_n\}_n=x$ so it must be bounded by $x$.Let's first construct the sequence $\{x_n\}_n$ of rational numbers in the invervall $(x-1,x)$.

For $n=1$ , by the density of $\mathbb{Q}$, there exists a rational number $r_1$ s.t: $$x-1<r_1<x.$$

For $n=2$, again by the density of $\mathbb{Q}$ we will have a rational number $r_2$ st: $$r_1<r_2<x.$$

...

For $n=N$ we will have: $$r_{N-1}<r_N<x.$$

We have shown that we can created a sequence $\{x_n\}_n$ of rational number that is strictly increasing and bounded by $x$.

Without loss of generality, we can created a sequence $\{y_n\}_n$ of irrational number that converge to $x$ knowing that $\mathbb{Q^c}$ is dense in $\mathbb{R}.$

Is my proof correct ? Any feedback is appreciated thank you.

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  • $\begingroup$ That's not enough. The sequence $1 - 1/n$ is bounded from above by $2$ but does not converge to it. $\endgroup$ – Santiago Jun 12 '16 at 15:06
  • $\begingroup$ So, what is your question??? $\endgroup$ – guestDiego Jun 12 '16 at 15:06
  • $\begingroup$ proof verification $\endgroup$ – Elina Jun 12 '16 at 15:16
  • $\begingroup$ @santiago, but the limit converge to 1 $\endgroup$ – Elina Jun 12 '16 at 15:23
  • $\begingroup$ ok i get it now my proof does not show that it converge. So i guess the best way is to proof by exemple, $\endgroup$ – Elina Jun 12 '16 at 18:43
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First let $x \in \mathbb{Q}$. Take the sequence $x_i=x- \frac{1}{i}$ for all $i$. Then $\left\{ x_i \right\}_i \subseteq \mathbb{Q}$ and this sequence clearly converges to $x$. For the other sequence, let $\left\{y_i \right\}_i \subseteq \mathbb{Q}^c$, $y_i= x- \frac{1}{\sqrt{i^2+1}}$. For $i \in \mathbb{N}$, $y_i \notin \mathbb{Q}$ and clearly $\lim y_i= x$. Both of these are strictly increasing.

Similar sequences can be constructed for $x \notin \mathbb{Q}$.

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  • $\begingroup$ thx, can you give any feedback on my proof please? thx you. $\endgroup$ – Elina Jun 12 '16 at 18:14
  • $\begingroup$ In your proof, you have not guaranteed convergence to $x$. If you can show, for instance, $\sup(x_n)=x$, then your proof is sufficient. This will be difficult with a sequence that is not explicitly constructed, like yours. $\endgroup$ – M10687 Jun 12 '16 at 18:21
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    $\begingroup$ Ok thank you, all the anwser here peoples have shown a proof by exemple that is why. $\endgroup$ – Elina Jun 12 '16 at 18:45
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Proof by example: $$x_n = \left(1+\frac{1}{n}\right)^n$$ $$y_n=e-\frac{1}{n}$$ where both sequences are strictly increasing and converge to the irrational number $e$.

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  • $\begingroup$ yeah your sequences does work . thank you. Please I would like some feedback about my proof, I'm not very confident about proof writing. thank you. $\endgroup$ – Elina Jun 12 '16 at 18:16
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Let $x_n = -1/n, y_n = -\sqrt 2/n.$

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  • $\begingroup$ thx very much , also i would like to know if my proof is correct. $\endgroup$ – Elina Jun 12 '16 at 18:14
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    $\begingroup$ Your proof as it stands is not complete as pointed out in the comments. You have simply created a sequence of rationals that are bounded above by x. That doesn't imply it converges to x. $\endgroup$ – zhw. Jun 12 '16 at 21:07

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