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Find limit of sequences $(x_n) = \frac{1}{n}, (y_n) = -\frac{1}{n}, (z_n) = (-1)^n$ in topological space $\tau = \{(a, +\infty), a\in \mathbb{R}\} $

For the first one the limit is A = $(-\infty, 0]$. For the second one the limit is B = $(-\infty, 0]$ . For the third one the limit is C = $(-\infty, -1]$

I think those are correct answers, but what are the answers if there is a diffrent topology $T = \{U \in \mathbb{R}; \mathbb{R}\setminus U <\aleph_0\}$? I think answer for all those sequences is $\mathbb{R}$

Matija
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k12
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  • What do you call limit of the series (do you mean sequences ?). – EDX Jan 18 '23 at 15:37
  • yes, i mean sequences, my mistake – k12 Jan 18 '23 at 15:38
  • And I don"t understand you call a limit of a squence a set like A. Could you precise ? – EDX Jan 18 '23 at 15:39
  • Yes, a set of points that satisfy this defintion of a limit: https://en.wikipedia.org/wiki/Limit_of_a_sequence#Topological_spaces – k12 Jan 18 '23 at 15:47
  • Can you write down clearly who is the topological space and who is the topology you are considering? I guess $\tau$ is the topology you want to give to $\mathbb R$, is that correct? Maybe I am wrong, but it doesn't seem to satisfy the definition of topology – Lorenzo Pompili Jan 18 '23 at 15:59
  • Topological space is pair $(\mathbb{R}, \tau)$ – k12 Jan 18 '23 at 16:04
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    Also, calling a set of points "the limit" of the sequence I suspect is a non standard thing to do, so you could write down explicitly what you mean by limit when you refer to a set instead of a single point – Lorenzo Pompili Jan 18 '23 at 16:05
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    I see. I fear that is not a topological space, as the union of open sets is not always open. Consider for instance the union of the sets $[1/n,+\infty)$ for positive, integer $n$. This is $(0,+\infty)$, which is not in $\tau$. Unless you mean that $\tau$ is the family of closed subsets – Lorenzo Pompili Jan 18 '23 at 16:07
  • You are absolutely correct, I made a mistake, the interval should be open on left side. What about the second topological space $(\mathbb{R}, T)$ ? – k12 Jan 18 '23 at 16:14
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    With that you mean the topology of all the sets whose complement is finite (you should also add the empty set)? Yes, the limit would always be $\mathbb R$, except for the third sequence for which the limit is the empty set – Lorenzo Pompili Jan 18 '23 at 16:32
  • In that case your answers are correct. Your description of the second topology doesn't make sense. Is it meant to be the cofinite topology on the reals? – Zoe Allen Jan 18 '23 at 16:32
  • No, because the set ${+1,-1}$ is finite, so the complement is open, so any point besides $\pm1$ are not limits. With a similar reasoning, you can exclude $\pm1$ – Lorenzo Pompili Jan 18 '23 at 18:53
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    @LorenzoPompili So in this defintion of limit, for x0 to not be a limit I just need to find one neighbourhood that doesn't satisfy the condition? – k12 Jan 18 '23 at 20:22

2 Answers2

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The answers for $\tau$ are correct. For the co-finite topology $T$, the answer can be found here.

Consider a sequence $(x_n)_n$. First assume that $|\{n:x_n=c\}|<\infty$ for all $c\in\mathbb R$. For $x\in\mathbb R$ and $U\in T$ with $x\in U$, so $|\mathbb R\setminus U|<\infty$. Thus, there exists $N$ such that $x_n\in U$ for all $n\ge N$. By definition, $x$ is a limit point and hence the limit is $\mathbb R$. This covers the first two examples.

For the last example, consider the remaining case that there exists a value $x^*$ which is taken infinitely many times. Further, assume there is a second value $x^\circ$ which is also taken infinitely many times. For $x\in\mathbb R\setminus\{x^*\}$ consider $U=\mathbb R\setminus\{x^*\}$, then for all $N$ there exists $n\ge N$ with $x_n=x^*\not\in U$ by definition, so $x$ is not a limit. By symmetry of $x^*$ and $x^\circ$ we obtain that $x^*$ is not a limit, so there are no limits here. This covers the last example.

For completeness, consider the case that there exists exactly one value $x^*$ that is taken infinitely many times. As before, we see that $\mathbb R\setminus\{x^*\}$ are not limit points. For an open set $U\in T$ with $x^*\in U$ we have $x_n\in U$ for all sufficiently large $n$ since the finitely many points $\mathbb R\setminus U$ are only visitied a finite number of times. This shows that $x^*$ is the unique limit in this case. An example sequence that shows this behavior would be $x_n=\mathbf{1}\{n\in 2\mathbb Z\}\frac{1}{n}$, with unique limit $0$.

Matija
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By $T$ you mean co-countable topology over the reals? Note that in this topology a sequence converges only if it is (eventually) constant: Assume that, for contradiction, $(x_n)$ is a non-constant sequence that converges to some $r\in R$. We always have some open set $U=R-{(x_n)}$ and $U\cup{r}$ is an open set containing $r$ , but not any other terms of $(x_n)$. Thus, $r$ cannot be a limit of the sequence , the sequences you define has no limits as they are not eventually constant.

Sun Y. U.
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