If $\ 0<\varepsilon<\vert x\vert,\ $ prove that only finitely many numbers of the form $\ \frac{(-1)^{k}}{2^{k}}\ $ are in $\ N(x;\varepsilon).$

Question

If $\ 0<\varepsilon<\vert x\vert,\ $ prove that only finitely many numbers of the form $\ \frac{(-1)^{k}}{2^{k}},\ k\in\mathbb{N},\ $ are in $\ N(x;\varepsilon).$

My proof is :

Let $x>0.$

Let $$S = \left\{ \left(-\frac{1}{2}\right)^k:\ k\in \mathbb{N}\right\}$$ Then $S$ has only one limit point $x_{0}=0$.

Suppose there are infinitely many points of $S$ in $N(x, \varepsilon)$. Let the set of these points $T$. Because T is a bounded, infinite set, by Bolzano-Weierstrass Thm, T has a limit point. Let one limit point of T be $x_{1}$. Because $T\subset S$, $x_{1}$ is also limit point of $S$. Thus $x_{1}=x_{0}=0$.

We have a contradiction because $x-\varepsilon >0$.

• I think this part is something uncertain. Can you explain?

Thus there are only finitely many points of $S$ in $N(x, \varepsilon)$.

Answer 1

It follows from the definition of $\Bbb R$ that any $r\in\Bbb R$ is less than some member of $\Bbb Z^+.$ And by induction on $n\in\Bbb Z^+ $ that $2^n>n$ for any $n\in \Bbb Z^+.$

Now $0<\varepsilon < |x|\implies N(x;\varepsilon)\cap N(0;|x|-\varepsilon)=\emptyset.$

So let $r=\dfrac {1}{|x|-\varepsilon}\,.$ We have $r>0.$ Take $n\in \Bbb Z^+$ with $n>r.$ Now $$n\le k\in\Bbb Z^+\implies 2^k>k\ge n>r\implies$$ $$ |(-1)^k2^{-k}|<1/r=|x|-\varepsilon\implies$$ $$(-1)^k2^{-k}\in N(0;|x|-\varepsilon)\implies$$ $$ (-1)^k2^{-k}\not\in N(x;\varepsilon).$$

Answer 2

I think that if you were to use $\begin{cases}\frac{(-1)^{k/2}}{2^{k/2}}&k\text{ even}\\x&k\text{ odd}\end{cases}$, then your "proof" would go through, but in this case, the conclusion is false, there are infinitely many points in the ball.

It read to me like you're assuming that the limit point of a sequence is unique, but that doesn't need to be true. So, the first place that I see a substantial error is when you conclude that $x_1=x_0=0$. It happens to be true in the current case, but you haven't proved that.

Answer 3

Here is a non-constructive proof in addition to @DanielWainfleet's constructive proof which I have to admit, is cooler in some ways than this.

Obviously the sequence $(-\frac{1}{2})^k$ goes to $0$ as $k\to\infty$.

WLOG, assume $x>0$. With $x>\epsilon>0$ we get $x-\epsilon>0$ which means $$\exists n_0\in\mathbb{N}:\forall k\geq n_0:|(-\frac{1}{2})^k|<x-\epsilon$$ which means $$\exists n_0\in\mathbb{N}:\forall k\geq n_0:(-\frac{1}{2})^k\notin N(x,\epsilon)$$ which means there are at most $n_0$ such numbers in the neighbourhood of x.

QED

Answer 4

I actually think your proof is fine, but since the result is obvious, the author of the question probably wants you to be more rigorous than you have been. I've put remarks and changes I would make in italics.

Let $x>0.$

Fine, but don't you then need to prove it for $\ x < 0\ $ also?

Let $$S = \left\{ \left(-\frac{1}{2}\right)^k:\ k\in \mathbb{N}\right\}$$

Then $S$ has only one limit point $x_{0}=0$

You need to give a proof of this, not just state it. [although this is arguable]

$$$$

Suppose there are infinitely many points of $S$ in $N(x, \varepsilon)$. Let the set of these points $T$. Because T is a bounded, infinite set, by Bolzano-Weierstrass Thm, T has a (at least one) limit point.

Let one limit point of T be $x_{1}$. Because $T\subset S$, $x_{1}$ is also limit point of $S$.

This is true, but you need to justify it, i.e. You need to prove the theorem that: if $x$ is a l.p of $X$ and $X\subset Y$ then $x$ is also a l.p. of $Y$.

Thus $x_{1}=x_{0}=0$.

We have a contradiction because $x-\varepsilon >0$.

You need to give more detail/be more precise as to what the contradiction is.

Thus there are only finitely many points of $S$ in $N(x, \varepsilon)$.

[Your proof is valid and sound...]