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Discuss the convergence of each of the following sequences in the spaces indicated.

a) $s_n=1+\frac{1}{n}$, in the space of real numbers with the absolute value metric.

The sequence will converge to point $x=1$ for $n$ large. Choose a length $p>0$ around $1$, the chosen $p$ will give us an open set with values in $(1-p,1+p)$ of the reals. Indeed, $s_n$ converges.

b) $s_n=(2,2)$, in the place $R^2$ with the metric $D_2$ or discrete metric.

This converges to itself because for $p$ being a small radius, all the terms tend toward the point $(2,2)$.

c) $s_n=(2,n)$, in the plane $R^2$ with metric $D_3=max\lbrace \lvert x_1-x_2 \rvert, \lvert y_1-y_2 \rvert \rbrace$.

This sequence will not converge because for any radius $p>0$,

max$\lbrace \lvert x_1-x_2 \rvert, \lvert y_1-y_2 \rvert \rbrace=\lvert y_1-y_2 \rvert>p$ as $n$ tends to infinity.

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  • $\begingroup$ To start with a), what does $p>0$ mean? $\endgroup$
    – Ian Coley
    Sep 26, 2013 at 23:00
  • $\begingroup$ Hopefully I have made it clear. $\endgroup$ Sep 26, 2013 at 23:10
  • $\begingroup$ I'm still confused what your problem is. For any small $p>0$, infinitely many $s_n$ are contained in $(1-p,1+p)$. $\endgroup$
    – Ian Coley
    Sep 26, 2013 at 23:15
  • $\begingroup$ Yes, considering 1+p>1. But, in order to have an open set we must have that all points are contained within the interval (1-p,1+p) which is false if $s_n=1$. That is what I am trying to say, now, it may be incorrect. $\endgroup$ Sep 26, 2013 at 23:24
  • $\begingroup$ The way I am seeing this is all points $y\in (1-p,1)$ are in the open set, but for any $z\in (1+p)$ is not. $\endgroup$ Sep 26, 2013 at 23:26

1 Answer 1

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I continue to be confused by your comments, so I'll just write an answer.

For a), we need to show that, for any $\varepsilon>0$, there exists $N\in\mathbb N$ such that $|s_n-1|<\varepsilon$ for all $n>N$. By the Archimedean property of reals, there exists $N\in\mathbb N$ such that $\frac1N<\varepsilon$, hence $\frac1n<\varepsilon$ for $n>N$. Then for $n>N$, $$ |1-s_n|<|1-(1+\varepsilon)|=\varepsilon. $$ Thus $s_n\to 1$ does converge.

Your conclusions for b) and c) are correct. You may also show c) does not converge because the sequence is not Cauchy.

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  • $\begingroup$ Thank you for a) come to think of it this is true because none of the terms of $s_n$ lie after 1. $\endgroup$ Sep 26, 2013 at 23:51

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