Say, $c$ with $c>0$, is an irrational number, what I need to prove is, any $\delta$-ngbhd of $c$ i.e. $V_{\delta}(c)$ contains a finite number of rational numbers.

I am not sure if the statement is correct or not.

Could anyone comment/provide a hint for the proof ?

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    $\begingroup$ The statement is false $\endgroup$ – dEmigOd Nov 14 '17 at 9:03
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    $\begingroup$ $\mathbb Q$ is dense in $\mathbb R$ hence each $\delta$ neighborhood of each value $c\in\mathbb R$ contains infinitely many rational and infinitely many irrational points if you talk about open sets wrt the standard topology on $\mathbb R$. $\endgroup$ – Mundron Schmidt Nov 14 '17 at 9:03
  • $\begingroup$ Are you sure you mean "finite" not "infinite"? $\endgroup$ – 5xum Nov 14 '17 at 9:05
  • $\begingroup$ Hint: Note that for $\sqrt{2},$ all but finitely many (hence, infinitely many) of the following infinitely many rational numbers belong to any specified $\delta$-neighborhood of $\sqrt{2}:$ $1.4,$ $1.41,$ $1.414,$ $1.4142,$ $1.41421,$ $1.414213,$ $1.4142135,$ $1.41421356,$ $\ldots$ $\endgroup$ – Dave L. Renfro Nov 14 '17 at 11:34

If you have $5$ cars, do you have $3$ cars?

By denseness, we know that the neighborhood indeed contains infinitely many rational numbers.

Does the neighborhood contains $5$ rational numbers?

I would view "the neighborhood containly only finitely many numbers" as a False statement.

But we can certainly find finitely many numbers in that neighborhood.


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