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So the question is:

Suppose $x_i$ and $y_i$ are infinite sequences of real numbers converging to x and y. Show that $(x_i + y_i)$ converges to $x+y$. Show that $x_iy_i$ converges to $xy$.

Here's what I tried for $x_iy_i$:
$|x_i - x| < \epsilon$ for all $\epsilon > 0$ there is an $n$ $\epsilon$ $N$ such that $n \le N$. and
$|y_i - y| < \epsilon$ for all $\epsilon > 0$ there is an $n$ $\epsilon$ $N$ such that $n \le N$.
$(x_iy_i) \to |x_i-x||y_i-y| < \epsilon$
$|x_iy_i - xy_i - x_iy + xy| < \epsilon$
but I can't seem to figure out how to make it so that $|x_iy_i - xy| < \epsilon$


Any hints or suggestions on what I should do to make it work? Thanks!

Zoë Soriano
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1 Answers1

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Hint: $|x_i+y_i-x-y|\le |x_i-x|+|y_i-y|$.

Another hint: $|x_i y_i-x y|\le |x_i y_i-x_i y+x_i y-x y| \le |x_i| |y_i- y|+|y||x_i -x |$.

TZakrevskiy
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  • Another hint: Use that a convergent series is bounded. – Marc Dec 13 '14 at 15:19
  • @Marc of course. But a sequence, not a series. – TZakrevskiy Dec 13 '14 at 15:19
  • @Marc I thought I did use the definition of convergence in my proof or was I wrong? $|x_i - x| < \epsilon$ for all $\epsilon > 0$ there is an $n$ $\epsilon$ $N$ such that $n \le N$.& thanks TZakrevskiy that's really helpful – Zoë Soriano Dec 13 '14 at 15:25