Suppose we are given the sequence $\{a_n\}_{n=1}^\infty$ defined by the rule $\forall n\in\mathbb{N}, a_n=n\sin \frac{\pi n}{4}$.
We can show that this is equivalent to:
$\forall n\in\mathbb{N},a_n= \begin{cases}
\frac{n}{\sqrt{2}} & n \equiv 1\pmod{8} \lor n \equiv 3\pmod{8} \\
n & n \equiv 2\pmod{8} \\
0 & n \equiv 0\pmod{8}\lor n \equiv 4\pmod{8} \\
-\frac{n}{\sqrt{2}} & n \equiv 5\pmod{8} \lor n \equiv 7\pmod{8} \\
-n & n\equiv 6 \pmod{8}
\end{cases}
$
Now the problem is to prove\show that around each number we take $L\in\mathbb{R}$ we can create\build a neighborhood such that:
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If $L$ does not equal to any element in the sequence $\{a_n\}_{n=1}^\infty$, Then we can build a neighborhood of $L$ such that no other element from the sequence will reside in it.
-
If $L$ equals some element in the sequence $\{a_n\}_{n=1}^\infty$, Then, If $L$ equals zero (as zero is an element in the sequence), Then we can build a neighborhood of $L$ that will contain an infinite number of elements from the sequence, And If $L$ does not equals zero, Then we can build a neighborhood of $L$ that will contain exactly one element from the sequence.
Translating the problem into logic/set-theory notation we get that we have to show:
-
If $\forall n\in\mathbb{N}, L\neq a_n$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=0$
-
If $\exists m\in\mathbb{N}, L= a_m$, Then,
If $L=0$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=\aleph_0$,
And If $L\neq 0$, Then $\exists \epsilon\in (0,\infty), |\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=1$
Note: The notation $N_\epsilon (L)$ denotes the neighborhood of $L$ with radius $\epsilon$, I.e. $N_\epsilon(L)=(L-\epsilon,L+\epsilon)$
My try for proving 1:
Since in this case it is given that $\forall n\in\mathbb{N}, L\neq a_n$ we can conclude that $\forall n\in\mathbb{N}, a_n-L\neq 0$ and thus $\forall n\in\mathbb{N}, | a_n-L|>0$, Now If we define the set $A=\{|a_n-L| | n\in\mathbb{N}\}$ we get that this set is a non-empty set of real numbers that is bounded below by $0$ and thus its infimum $\inf(A)=\inf\limits_{n\in\mathbb{N}} |a_n-L|$ exists and satisfies $\inf\limits_{n\in\mathbb{N}}|a_n-L|\geq 0$ (as $0$ is a lower bound of $A$ and thus must be less than or equal to its infimum), Now if we just show that $\inf\limits_{n\in\mathbb{N}}|a_n-L|>0$ we could choose $\epsilon=\inf\limits_{n\in\mathbb{N}}|a_n-L|\in (0,\infty)$ and we would get that $\forall n\in\mathbb{N}, a_n\notin N_\epsilon(L)$ because if we suppose by contradiction that $\exists n\in\mathbb{N}, a_n\in N_\epsilon(L)$, Then we would get by the fact that $|a_n-L|\in A$ and by definition of infimum that $|a_n-L|\geq \inf(A)=\inf\limits_{n\in\mathbb{N}}|a_n-L|=\epsilon$ which contradicts the fact that $a_n\in N_\epsilon(L)$ (which is equivalent to $|a_n-L|<\epsilon$), Thus it must be the case that $\forall n\in\mathbb{N},a_n\notin N_\epsilon(L)$ and we can conclude that $\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}=\emptyset$ and thus $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=0$ as was to be shown.
My try for proving 2:
Since in this case it is given that $\exists m\in\mathbb{N}, L = a_m$,
And since $0$ is in the image of $\{a_n\}_{n=1}^\infty$ (i.e. $0$ is an element in the set $\{a_n|n\in\mathbb{N}\}$) we get that there are two possibilities: $L=0$ or $L\neq 0$
If $L=0$, Any $\epsilon$ we choose from the set $(0,\infty)$ will work, As we can define the set $T=\{n\in\mathbb{N}|n\equiv 0 \pmod{8}\}$ and it is clear that for this set we have $\forall n\in T, a_n\in N_\epsilon(L)$ [because $\forall n\in T, a_n=0$ and because $0\in N_\epsilon(0)=N_\epsilon(L)$], And therefore we get that $T\subseteq \{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}\subseteq \mathbb{N}$, But since $|T| = |\mathbb{N}| =\aleph_0$ we get that it must be the case that $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=\aleph_0$ as was to be shown.
Now if $L\neq 0$, We can define the set $A=\{|a_n-L| | m\neq n\in\mathbb{N}\}$, And we get that this set is a non-empty set of real numbers that is bounded below by $0$ and thus its infimum $\inf(A)=\inf\limits_{m\neq n\in\mathbb{N}} |a_n-L|$ exists and satisfies $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|\geq 0$ (as $0$ is a lower bound of $A$ and thus must be less than or equal to its infimum), Now if we just show that $\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|>0$ we could choose $\epsilon=\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|\in (0,\infty)$ and we would get that $\forall m\neq n\in\mathbb{N}, a_n\notin N_\epsilon(L)$ because if we suppose by contradiction that $\exists m\neq n\in\mathbb{N}, a_n\in N_\epsilon(L)$, Then we would get by the fact that $|a_n-L|\in A$ and by definition of infimum that $|a_n-L|\geq \inf(A)=\inf\limits_{m\neq n\in\mathbb{N}}|a_n-L|=\epsilon$ which contradicts the fact that $a_n\in N_\epsilon(L)$ (which is equivalent to $|a_n-L|<\epsilon$), Thus it must be the case that $\forall m\neq n\in\mathbb{N},a_n\notin N_\epsilon(L)$ and we can conclude that $\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}=\{m\}$ (because $L=a_m$) and thus $|\{n\in\mathbb{N}|a_n\in N_\epsilon(L)\}|=1$ as was to be shown.
Thanks for any hint\help on how to prove that those infimums are indeed positive….
Best Answer
Hints:
If $L$ is not an element of the sequence then there exists unique $n \in \mathbb{Z}$ such that $n < L < n+1$ and unique $m \in \mathbb{Z}$ such that $\frac{m}{\sqrt2} < L < \frac{m+1}{\sqrt2}$.
Therefore let $\varepsilon = \min\left\{\left|L-(n+1 )\right|, \left|L-n\right|, \left|L-\frac{m+1}{\sqrt2}\right|, \left|L - \frac{m}{\sqrt2}\right|\right\}$ and consider the neighbourhood $(L-\varepsilon, L + \varepsilon)$.
If $L = 0$ then any neighbourhood of $0$ will do, because $0$ appears infinitely many times in the sequence.
If $L$ appears in the sequence but $L \ne 0$, then:
if $L \in \mathbb{Z}$ then there exists a unique $m \in \mathbb{Z}$ such that $\frac{m}{\sqrt2} < L < \frac{m+1}{\sqrt2}$. Let $\varepsilon = \min\left\{1, \left|L-\frac{m+1}{\sqrt2}\right|, \left|L - \frac{m}{\sqrt2}\right|\right\}$.
if $L = \frac{n}{\sqrt2}$ for some $n\in \mathbb{Z}$ then there exists a unique $m \in \mathbb{Z}$ such that $m < L < m+1$. Let $\varepsilon = \min\left\{\frac1{\sqrt2}, \left|L-(m+1)\right|, \left|L - m\right|\right\}$.
and consider the neighbourhood $(L-\varepsilon, L + \varepsilon)$.