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What's the boundary of a sequence of integers?

I'm trying to learn about topology, particularly boundaries and open and closed sets. I think I correctly grasp the idea that some point in the real line can be a limit point of a sequence and therefore such a sequence may not contain its boundary.

But I'm struggling to get to grips with the boundary of a sequence of integers in $\mathbb{N}$. It would seem $\mathbb{N}$ has the discrete topology. If we take the sequence $3,4,5,6$ then is its boundary in $\mathbb{N}$, $\{2,7\}$ or $\{3,6\}$? It would seem we can define this set as an open or closed set with two different boundaries, but the set itself is unchanged so it makes no sense to say the set itself is open or closed as it's either, and we can choose at will what the boundary of the same set is. Am I being daft or overlooking something obvious?

it's a hire car baby
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    The boundary operation is defined for subsets of a topological space. You don't mention it, but likely they intended you to compute the boundary of $\mathbb{N}$ as a subset of $\mathbb{R}$ with the usual topology. The boundary is the closure minus the interior. The set of natural numbers is closed and you cannot fit any open interval inside so its interior is empty. Therefore the boundary is all of $\mathbb{N}$ minus the empty set, resulting in all of $\mathbb{N}$ – Peyton Aug 02 '17 at 13:19
  • There is no "They". I can get on board with the possibility that the sequence is its own boundary in $\mathbb{R}$ but I'm interested in the discrete topology. – it's a hire car baby Aug 02 '17 at 13:20
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    They is you. The induced topology on $\mathbb{N}$ as a subset of $\mathbb{R}$ is the discrete topology. In any case, boundary of a set is an operation that requires a superset, it is relative to a topological space that contains the set. It has to be mentioned as the answer depends on it. So, who is your superset? – Peyton Aug 02 '17 at 13:24
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    If $X=\mathbb{N}$ is your topological space, and you want to compute boundary of its subset $\mathbb{N}\subset X$. Then boundary of $\mathbb{N}$ is the $\emptyset$. This is independent of who $X$ is and independent of its topology. Compare now with the computation above. If $X=\mathbb{R}$ and we are computing the boundary of $\mathbb{N}\subset X$. Then the boundary is all of $\mathbb{N}$. This shows both that the answer depends on the superset and its topology, and not on the induced topology on the subset. In both cases the subset $\mathbb{N}$ has the discrete topology induced. – Peyton Aug 02 '17 at 13:33
  • @Peyton Are you saying that if we take the boundary in $\mathbb{R}$ with the usual topology then $\partial{3,4,5,6}=\mathbb{N}$? – it's a hire car baby Aug 02 '17 at 16:47
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    No, the boundary of ${3,4,5,6}$ as a subset of the topological space $\mathbb{R}$ endowed with its usual topology, is equal to ${3,4,5,6}$. – Hellen Aug 02 '17 at 17:02
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    But to be clear of the relevance of $\mathbb{R}$ in the comment above, if you consider taking the boundary of ${3,4,5,6}$ as a subset of the topological space $\mathbb{N}$ endowed with the discrete topology, then the boundary of ${3,4,5,6}$ is the empty set. – Hellen Aug 02 '17 at 17:04
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    "It makes no sense to say the set is open or closed as it's either." WRONG. A set can be both open and closed. In any space $X$ the set $X$ and the empty set are open-and-closed sets. In a discrete space every subset is open-and-closed. In the space $[0,1] \cup [2,3]$ the set $[0,1]$ is open-and-closed – DanielWainfleet Aug 02 '17 at 18:14
  • @DanielWainfleet thanks. I can see that now. It had not occurred to me when I wrote this question that the topology of $\mathbb{N}$ was viewed so fundamentally as the set of points arranged in the spacing we see on the real number line. I imagined the space where we removed the rest of the real numbers to be abstracted away and two integers to be in the neighbourhood of each other, rather than separated by the void the non-integers left. – it's a hire car baby Aug 02 '17 at 18:49
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    Q: Is it actually possible to define a topology on $\mathbb{N}$ such that (1) all neighborhoods contain more than one point, (2) all points have a finite neighborhood, and (3) all points are topologically distinguishable? – Michael L. Aug 02 '17 at 19:14
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    Another Q: Is it possible to define a topology on $\mathbb{N}$ such that (1) all neighborhoods contain more than one point and (2) $\mathbb{N}$ with this topology is Hausdorff? – Michael L. Aug 02 '17 at 19:37
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    @MichaelLee For your second question, sure - make $\mathbb{N}$ homeomorphic to $\mathbb{Q}$ with the usual topology. For your second question, the answer is no: iteratively apply topological distinguishability to thin an arbitrary finite neighborhood to a single-element neighborhood. (If $x\not= y$ are in some finite open set $U$, let $V$ be an open set distinguishing $x$ and $y$ - that is, so that $\vert V\cap {x, y}\vert=1$ - and note that $U\cap V$ is still open and has strictly fewer points but also has at least one point.) – Noah Schweber Aug 02 '17 at 19:39
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    @MichaelLee. By contradiction: Let $n_0$ be the least number of members of a nbhd. Let $U_0$ be an $n_0$-member nbhd of $p_0.$ If $p\in W\subsetneqq U_0$ then $W$ is not open by minimality of $n_0$, but there exists open $V$ with $p_0\in V\subset U_0,$ so $U_0$ is open. There exists $p_1\in U_0$ \ ${p_0}.$ If $V\supset U_0$ for every open $V$ containing $p_1$ then $p_0, p_1$ would have the same set of nbhds. So there exists open $U_1\ne U_0 $ with $p_1\in U_1$. But $ U_1\cap U_0$ is a nbhd of $p_1$ with less than $n_0$ members, contrary to the minimality of $n_0.$ That was fun. – DanielWainfleet Aug 02 '17 at 19:46
  • @NoahSchweber. Your reply to Michael Lee appeared while I was trying to pare down my reply to <500 characters. – DanielWainfleet Aug 02 '17 at 19:47
  • @MichaelLee. Erratum in my previous comment: It should say "So there exists open $ U_1 \not \supset U_0$ with $p_1\in U_1,$". .... I ran out of editing time. – DanielWainfleet Aug 02 '17 at 19:55
  • @NoahSchweber does it change the answer to Michael's first question if we represent each integer with a pair of points, one on its downside and one representing it's upside? This possibility makes me think of the parity problem for some reason. – it's a hire car baby Aug 03 '17 at 03:24

3 Answers3

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A point $a$ of a topological space $X$ is called a boundary point of $A\subseteq X$ if for each nbhd $U$ of $a$ we have $U\cap A,U\cap (X\setminus A)\neq \emptyset$. So as already mentioned in comments you need to know what the underlying topology is. You can give $\mathbb{N}$ a discrete topology and consider it as a subset of itself, and in that case you can see that $\partial\mathbb{N}=\emptyset$. If $\mathbb{N}$ is considered as a subset of $\mathbb{R}$ with the usual topology then $\partial\mathbb{N=N}$. For take any point $n\in\mathbb{N}$ and a nbhd $U$ of $n$. Then $U\cap \mathbb{N}\neq\emptyset$ and $U\cap (\mathbb{R\setminus N})\neq\emptyset.$ So the point is that the boundary of a set depends on the open sets or rather the underlying topology. Does this help?

Peyton
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Janitha357
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For a subset $S$ of a space $X,$ the boundary $\partial S$ (also written $Fr(S),$ as $Fr$ stands for Frontier) is $$\partial S=\overline S \cap \overline {X \setminus S}.$$ Sometimes we write $\partial_X S$ because it depends on the space $X.$

If we take $X=\mathbb N$ where the topology on $\mathbb N$ is its subspace topology as a subspace of $\mathbb R$ (with the usual topology on $\mathbb R),$ then $X$ is a discrete space: Every subset of $X$ is closed. So if $S\subset \mathbb N$ then $\partial_{\mathbb N}S= \overline S \cap \overline {X \setminus S}=S \cap (X$ \ $S)=\phi.$

If we take $X=\mathbb R$ with the usual topology, then $\mathbb N$ is a closed discrete subspace of $X$ so every $S\subset \mathbb N$ is closed in $X.$.... And if $S\subset \mathbb N$ then $\overline {X \setminus S}\supset$ $ \overline {X \setminus \mathbb N}=\overline {\mathbb R \setminus \mathbb N}=\mathbb R=X,$ so $\overline {X \setminus S}=X.$ Hence for any $S\subset \mathbb N$ we have $\partial_{\mathbb R}S=\overline S \cap \overline {X \setminus S}=S \cap X=S.$

Michael Hardy
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DanielWainfleet
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  • Thanks. I'm intrigued by the apparently paradoxical convention to always view the topology on $\mathbb{N}$ as its subspace topology as a subspace of $\mathbb{R}$ rather than to derive its topology in total isolation from any other set of numbers. – it's a hire car baby Aug 02 '17 at 18:21
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    $\mathbb N$ with the discrete topology is a familiar example of a countably infinite discrete space, and it also happens to be a subspace of another familiar space, $\mathbb R $. Sometimes a preference for one of two or more equivalent definitions is a matter of taste. On the other hand there is a considerable list of properties, each equivalent to the topological def'n of "continuous function", and many of them are frequently-used "tools" . – DanielWainfleet Aug 02 '17 at 18:43
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If $\mathbb N $ has the discrete topology, the convergent sequences are eventually constant. ..

Your example is of a finite list of natural numbers . As a set, $\{3,4,5,6\} $ would have boundary $\emptyset $.

Sequences have limits, or don't converge, as the case may be. ..

The boundary of a set is the set of limit points, that is, points such that there is a sequence of points in the set converging to them, intersected with the closure of its complement. Alternatively, it is the closure minus the interior.

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    This question has got $4$ answer in a row and all $4$ are wrong. I am impressed. – Peyton Aug 02 '17 at 13:42
  • You're right. I have tried to correct it. .. –  Aug 02 '17 at 13:51
  • @Peyton, I would say that both answers os correct but that your comments are not. – Lehs Aug 02 '17 at 13:54
  • @Lehs Then you haven't even read the link that you yourself posted in your question. Your answer is wrong because you haven't understood that the boundary operator is a function from the power set of a topological space to the same power set. I can leave the topology of $\mathbb{N}$ fixed and change the space that contains it and change what the boundary is. Therefore the topology of the set of which you are computing the boundary of is irrelevant. The topology of the space that contains it is what determines the boundary – Peyton Aug 02 '17 at 13:55
  • @Peyton: I've red it and I never wrote that the closure wasn't a function $cl:\mathcal P(X)\to\mathcal P(X)$. – Lehs Aug 02 '17 at 14:02
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    @Lehs You wrote "Therefore it depends on the topology selected for the set of integers". That is wrong. – Peyton Aug 02 '17 at 14:02
  • Actually, I don't really see why it shouldn't depend on the topology selected. Different topologies would induce different Kuratowski closure operators, which doesn't contradict the closure axioms at all. For example, we can choose topologies on $\mathbb{N}$ such that a singleton is closed, and we can choose topologies on $\mathbb{N}$ such that a singleton is not closed. Obviously, the closures of ${1}$ in these two topologies will not be the same. – Michael L. Aug 03 '17 at 21:07
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    Could it be that there is a misunderstanding here? Michael Lee and Lehs want to tell the OP that endowing $\mathbb{N}$ with different topologies produces different closures. Peyton's point is that if, more generally, you interpret $\mathbb{N}$ as a subspace of some topological space, you can get different closures even if the subspace topology on $\mathbb{N}$ is the same. Right? – Torsten Schoeneberg Aug 08 '17 at 17:33