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I came across the following problem during the course of my study of real analysis:

Prove that $(x_n)$ is a null sequence iff $(x_{n}^{2})$ is null.

For all $\epsilon>0$, $|x_{n}| \leq \epsilon$ for $n > N_1$. Let $N_2 = \text{ceiling}(\sqrt{N_1})$. Then $(x_{n}^{2}) \leq \epsilon$ for $n > N_2$. If $(x_{n}^{2})$ is null then $|x_{n}^{2}| \leq \epsilon$ for $n>N$. Let $N_3 = N^2$. Then $|x_n| \leq \epsilon$ for $n> N_3$.

Is this correct? In general, we could say $(x_{n})$ is null iff $(x_{n}^{n})$ is null?

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+1 for bravely showing what you had tried. See Gortaur's answer for your first question, Shai Covo's for the last question. – Jyrki Lahtonen Jun 23 '11 at 13:55
+1 for asking a good question. – Patrick Da Silva Aug 8 '11 at 13:00
up vote 4 down vote accepted

That's not quite correct since you don't know the connection between $n$ and $x_n$ - hence taking square roots of $n$ are meaningless. What you can do is the following:

  1. $x_n\to 0$, hence for all $\epsilon>0$ exists $N$ such that $|x_n|<\sqrt{\epsilon}$ if $n\geq N$, but then it means that for all $\epsilon>0$ exists $N$ such that $|x_n|^2<\epsilon$ if $n\geq N$.

  2. $x_n^2\to 0$, hence for all $\epsilon>0$ exists $N$ such that $|x_n|^2<\epsilon^2$ if $n\geq N$, but then it means that for all $\epsilon>0$ exists $N$ such that $|x_n|<\epsilon$ if $n\geq N$.

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Concerning the last question, it is true that, for any fixed positive integer $k$, $x_n \to 0$ if and only if $x_n^k \to 0$. Also, $x_n \to 0$ obviously implies that $x_n^n \to 0$. On the other hand, $x_n^n \to 0$ does not imply $x_n \to 0$ (take for example $x_n = 1/2$ $\forall n$).

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You could use the following fact:

If a function $f:X\to Y$ between two topological spaces is continuous and $x_n\to x$, then $f(x_n)\to f(x)$.

(In case you do not have learned it in this generality, you might at least know that this is true for real functions or for functions between metric spaces. In fact, in the case of real functions the above condition is equivalent to continuity.)

You can obtain your first claim by applying the fact to the continuous functions:

$f: \mathbb R\to\mathbb R$, $f(x)=x^2$ (one implication)

$f: \langle 0,\infty)\to \mathbb R$, $f(x)=\sqrt{x}$ (reverse implication)

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Clearly the OP is not this much acknowledged in mathematics... topology is waaaay above beginning real analysis :P but you had a cool reflex :) +1 – Patrick Da Silva Aug 8 '11 at 13:00
I corrected two typos, I hope you did not mind :-) – Asaf Karagila Aug 8 '11 at 13:04
Thanks @Asaf... – Martin Sleziak Aug 8 '11 at 13:07
@Patrick I believe this result - for real functions - is quite often included in the first course in analysis. (Under the name Heine definition of continuity.) – Martin Sleziak Aug 8 '11 at 18:04

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