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I have a problems regarding finding limit points:

(1) $x_{2k} = a, x_{2k+1}=b$

(2) $x_1=1,\ \ x_2=2,\ \ x_k=k,\ \ x_{k+1}=1$

(3) 1,2,1,2,3,1,2,3,4,1...

My question is: How do I describe my solution rigorously and is my approach for rigorous proof of finding limit points good enough? What I do is the following:

We know that a limit point contains infinitely many members of the sequence in every open intervals of it.

(1) Let $\epsilon>0, \delta >0$ where $\epsilon$ and $\delta$ are real numbers. We define the interval $(a-\epsilon, a+\delta)$. We see that there are infinitely many members of the sequence in the interval since every $2k^{th}$ member is 2, and 2k goes to infinity. Therefore a is a limit point of the sequence. The proof for b being a limit point is the same.

(2) We define inductively the intervals $(n-\alpha_{k+n}, n+\beta_{k+n})$ where $\alpha_k, \beta_k$ are arbitrary real numbers. We see that for every $(k+n)^{th}$ member equals n, for n being positive natural number and that k+n goes to infinity, therefore there are infinitely many members equaling n in the $(k+n)^{th}$ interval. Therefore, every n..n+k number is a limit point of the sequence.

(3) Well, here I had an idea but I couldn't write it down very formally: Let's define the following sequence: $$ n_0=1 \\ n_{k+1}=n_k+k+1 $$ $n_k$ defines the number i, for $n_0=i$ so that I can show that every i occurs infinitely many times in an arbitrary interval of i.

My seek of knowledge on this question is solely for the purpose of rigorous and formal writing of a solution in the domain of the problem.

Thanks in advance!

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I will give rigorous answers for all the above questions. I hope this is satisfactory.

A limit point of the sequence is a point such that every neighbourhood of this point contains infinitely many points from the given sequence.

A neighbourhood of a point is defined as a set which contains an open set that contains that point.

1) $x_{2k} = a, x_{2k+1} = b$.

Here, I claim that $a$ and $b$ are the only limit points of the sequence.

To see this, let $U$ be any neighbourhood of $a$. Note that since for all $k$, $x_{2k} = a$ is in $U$ (by definition of $U$), $U$ contains infinitely many terms of the sequence $x_{k}$. Hence, $a$ is a limit point. A similar argument follows for $b$. Suppose $c$ is a point different from $a$ and $b$, then let $D = \min\{|c-a|,|c-b|\}$ of course, $D > 0$. Note that if we let $V = \{ y | |y - c| < \frac D2\}$, then $V$ is a neighbourhood of $c$ which does not contain any points of the sequence $x_k$. Hence, $c$ is not a limit point of the sequence $x_k$, implying that it's only limit points are $a$ and $b$.

2) $x_{1} = 1, x_2 = 2...x_k=k,x_{k+1} = 1, ...$

Here, I claim that the only limit points of the sequence are $1,2,...,k$.

Again, Let $U$ be any neighbourhood of $1$. Note that for all $n \in \mathbb N$, $x_{nk + 1} = 1 \in U$ by definition of $U$. Hence, it follows that $U$ contains infinitely many terms of the sequence $x_k$. Finally, this means that $1$ is a limit point of the sequence. Similarly, we can show that $2,...,k$ are also limit points of the sequence.

Let $c \notin \{1,2,...,k\}$, and let $D = \min\{|c-1|,|c-2|,...,|c-k|\}$. Again, $D > 0$. So if we let $V = \{ y | |y-c|<\frac D2\}$, then $V$ contains no elements of $x_k$. Hence, $c$ is not a limit point of $x_k$, giving the required conclusion.

3) $1,2,1,2,3,1,2,3,4,1,2,3,4,5,...$ is not a well defined sequence like the others. We just have a feel for the way it progresses.

We need a rigorous way of describing this sequence. To do this, note the positions that $1$ is coming at : $1,3,6,10,15,...$ this matches the sequence $\frac{n(n+1)}{2}$. The positions $2$ is coming at are always after $1$ : $\frac{n(n+1)}{2} + 1$. The positions $3$ are coming at are always after $2$, but only when $n \geq 2$ : $\frac{n(n+1)}{2} +2$ , but only for $n \geq 2$. This fits : $5,8,12$ etc.

With this in mind, we come up with a way of describing $x_k$, which goes like $1,2,1,2,3,1,2,3,4,...$ as follows:

Let $k \in \mathbb N$. Since the sequence $\frac{n(n+1)}2$ is a strictly increasing sequence with $n$, there exists $n$ such that $\frac{n(n+1)}2 \leq k < \frac{(n+1)(n+2)}2$. Define $x_k = k - \frac{n(n+1)}2 + 1$.

Does this work? Let $k=1$. Then, $n=1$ fits the bill, and $x_k = 1 - 1 + 1 = 1$.

Similarly, if $k=10$, then $10 \leq k < 15$, so $n = 4$ works, and $x_{10} = 10-10+1 = 1$ is correct.

You can confirm that this is the sequence we are looking for : it does go like $1,2,1,2,3,1,2,3,4,...$

Having done this, I claim that $1,2,3,...$ are all limit points of the sequence.

Consider any $m \in \mathbb N$. Let $U$ be a neighbourhood of $m$.

I leave you to prove a simple lemma:

For all $k \geq 2m$, $\frac{k(k+1)}{2} < \frac{k(k+1)}2 +(m-1) < \frac{(k+1)(k+2)}2$.

Having done this, consider , for $k \geq 2m$, the element $x_{\left(\frac{k(k+1)}2 +(m-1)\right)}$. By definition, it is equal to $\frac{k(k+1)}2 +(m-1) - \frac{k(k+1)}{2} + 1 = m-1+1 = m$. Since this happens for all $k \geq 2m$, we have that there are infinitely many terms of the sequence $x_k$, which are equal to $m$, and hence in the neighbourhood $U$. It follows that $m$ is a limit point of the sequence. Since $m$ is an arbitrary natural numbers, it follows that all natural numbers are limit points of the sequence.

If $c$ is not a natural number, then consider it's difference from the natural numbers,take the minimum (which exists) and then construct a neighbourhood $V$ which doesn't have any of the points $x_k$ (same as for the other two). Hence, we have an exact description of the limit points.

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