Real Analysis – No Sequence with Infinitely Many Zeros Converges to a Non-Zero Number

Question

I've already finished what I think is a valid proof by contradiction, but I would like to know if it's correct.

Let $(a_n)$ be a sequence with infinetly many $0$s, such that $\lim(a_n) = a \ , a\neq0$. Then $\forall\epsilon>0,\exists N\in \mathbb{N}$ such that $|a_n-a| < \epsilon \ ,\forall n>N$.

Let's choose an epsilon such that $0<\epsilon<|a|$. Knowing that $(a_n)$ has infinitely many $0$s, then $\exists n>N$ such that $a_n=0$. That way, for that particular $n$, we have that $|0-a|=|a|<|a|$, which is absurd.We can now conclude that only one of the following happens: $(a_n)$ doesn't converge to a non-zero number or $(a_n)$ has a finite number of $0$s.
Is this proof correct?

Before getting to this result I was trying to achieve that $\forall n > N$ we necessarily have that $a_n > 0 \ \lor \ a_n < 0$, but I was unable to do it. Is it possible to achieve something like this?

Answer 1

Let $(a_n)$ be a sequence with infinetly many $0$s, such that $lim(a_n) = a \ , a\neq0$. Then $\forall\epsilon>0,\exists N\in \mathbb{N}$ such that $|a_n-a| < \epsilon \ ,\forall n>N$.

Let's choose an epsilon such that $0<\epsilon<|a|$. Knowing that $(a_n)$ has infinitely many $0$s, then $\exists n>N$ such that $a_n=0$. That way, for that particular $n$, we have that $|0-a|=|a|<|a|$, which is absurd.We can now conclude that only one of the following happens: $(a_n)$ doesn't converge to a non-zero number or $(a_n)$ has a finite number of $0$s.

• Note the spelling of "infinitely".
• Limits should be typeset with \lim.
• Whenever possible, you should allow the typesetting engine (MathJax, LaTeX, whatever) to do its job. In several places, you have forced the typesetting engine to insert spaces in a manner that I, personally, find unaesthetic. Also, note that there should be spaces after the punctuation at the end of a sentence.
• What is $N$? The way the definition of convergence works, first $\varepsilon$ is chosen, and then a value of $N$ (which depends on $\varepsilon$) can be chosen.
• As a general rule (though this is a matter of taste), I prefer to use more words and less notation, unless the notation really helps to make the point, and I don't like to put two different notational phrases right next to each other (as I find that this makes things harder to read.
• Again, as a matter of opinion, good mathematical writing should be written in the imperative. I don't like the first-person plural (i.e. the mathematical "we"), though people do use it. (I find it affected.)

On a scale of "Needs Improvement" / "Satisfactory" / "Excellent", I would give this proof a "satisfactory". The writing is a little sloppy, and the ambiguity about the selection of $N$ is actually something of a real problem.

Taking the general idea and structure of your proof (which is basically correct), I would write something like the following:

Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that $\lim_{n\to\infty} a_n = a0$ and $a_n = 0$ for infinitely many $n$. Then for all $\varepsilon > 0$, there exists $N\in\mathbb{N}$ such that $n > N$ implies that $|a_n - a| < \varepsilon$.

For contradiction, suppose that $a \ne 0$, and fix some $\varepsilon$ such that $0 < \varepsilon < |a|$ and choose some $N \in\mathbb{N}$ so large that $n > N$ implies that $|a_n - a| < \varepsilon$. As the sequence contains infinitely many zeros, there exists some $n > N$ such that $a_n = 0$. For that $n$ $$ \varepsilon > |a_n - a| = |0 - a| = |a|. $$ But $\varepsilon$ was chosen so that $|a|>\varepsilon$, which is a contradiction. Therefore $a = 0$.