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I'm reading Baby Rudin at the moment and it claims something remarkable. Consider the sequence

$$ x_n=\frac{1}{n}.$$

The book claims that this converges to zero in the reals:

$$\lim_{n\to\infty} x_n=0.$$

It also claims that it does not converge like this in the positive reals. Why is this?

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    $\begingroup$ Where does it claim that? $\endgroup$ Jul 6, 2014 at 18:45
  • $\begingroup$ Damn, sorry, I seem to have misread it. It claims something else which I cannot understand. Will edit the question. $\endgroup$
    – user161959
    Jul 6, 2014 at 18:45
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    $\begingroup$ "Positive reals" means $(0,\infty)$. The point to which the sequence would converge does not belong to the space. $\endgroup$ Jul 6, 2014 at 18:47
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    $\begingroup$ Simply because $0$ is not positive. It's just a technicality: to converge in a particular space, the point of convergence also has to be in the space. $\endgroup$
    – angryavian
    Jul 6, 2014 at 18:47
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    $\begingroup$ This is by definition, more or less. $0$ is usually not taken to be a positive or a negative number. So the sequence is Cauchy in the positive reals but the value it converges to is not in the positive reals. $\endgroup$ Jul 6, 2014 at 18:47

2 Answers 2

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Because $0$ is not a positive real!

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the definition of the rudin book: "A sequence $\{p_n\}$ in a metric space $X$ is said to converge if there is a point $**p \in X**$ with the following property...$$\lim_{n\to\infty }\{p_n\}=p$$

how $0 \notin X=(0,\infty)$ then don't exist, a such $p\in(0,\infty)$

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