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So I'm currently studying from Rudin's Principles of mathematical analysis or colloquially "Baby Rudin" and have stumbled into the second chapter namely basic topology. He lists some sets and states whether or not they are bounded, open, closed or perfect. My question com3es from the fact that he calls the set of all integers closed

By the text a set is closed if every limit point is an element of the set itself. Naturally I understand the only limit points of the Integers to be $\infty$ and $-\infty$... however I assumed that the Integers don't contain either of these elements so I reasoned that the Integers were not closed

Could someone explain why this reasoning is wrong? I presume I'm misunderstanding something...

As a corollary question I was wondering which sets contain $\infty$ and/or $-\infty$

Continuing the story of my study I then assumed that I was wrong and that the Integers do in fact contain $\infty$ and $-\infty$ ( considering the set of complex and real numbers are as well considered closed I assumed that $\infty$ and $-\infty$ are elements of all these sets ) but then Rudin again talks about the set $S = \left\{\frac{1}{n} | \, n \in \mathbb{N} \right\} $ but says that $0$ is not an element (but is obviously a limit point)... I guess the confusion I have comes from the fact that I then assumed that $\infty$ is an element of the natural numbers and earlier he defines $\frac{1}{\infty} = 0$ so then $0$ should be in the set...

Where is my thinking going awry?

Thanks in advanced!

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Being "closed" is defined for a subset of another topology/metric space. The integers are closed as a subset of the real numbers. $\pm\infty$ are not in the real numbers. The integers are not closed in the topology of the $\mathbb R\cup\{+\infty,-\infty\}$. –  Thomas Andrews Mar 29 '13 at 16:58

3 Answers 3

up vote 6 down vote accepted

The integers are closed in $\Bbb R$, the space of real numbers; $\infty$ and $-\infty$ are not in that space and therefore are not relevant. Judging by a quick look at my second edition, he has not at that point talked about $\pm\infty$ or the extended real numbers at all.

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To elaborate slightly, if you apply the strict definition of 'limit point' to the natural numbers, you'll find that $\infty,-\infty$ don't fit it. You would need for every $\varepsilon>0$ the existence of some integer $n$ such that $|\infty-n|<\varepsilon$. But that's absurd since $\infty-n=\infty$ by its definition. –  Ian Coley Mar 29 '13 at 16:56
    
then how is it possible to find the limit of a sequence unless $\infty$ is in fact a limit point? –  DanZimm Mar 29 '13 at 16:57
    
@Dan: Many sequences of real numbers don’t have limits. For example, the sequence $\langle(-1)^n:n\in\Bbb N\rangle$ has no limit. The sequence $\langle n:n\in\Bbb N\rangle$ has no limit in $\Bbb R$; when we write that the limit is $\infty$, we either are using that statement as an abbreviation for the statement that for each $r\in\Bbb R$ there is an $m\in\Bbb N$ such that $x_n>r$ for each $n\ge m$, or are working in the so-called extended reals, which do include $-\infty$ and $\infty$. –  Brian M. Scott Mar 29 '13 at 17:00
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By applying the definition. $\lim_{n\to\infty}$ is well-defined without needing to know what $\infty$ is - think of it as notation, not that $\infty$ is an actual number. (Later, there are ways to formalize it so that $\lim_{n\to\infty}$ fits into the other definitions of limit, but that's probably too much to get to if you are just starting Rudin.) @DanZimm –  Thomas Andrews Mar 29 '13 at 17:00
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@Dan: You’re welcome! I think that topology is great fun, but I’m a topologist, so I might be a wee bit prejudiced. –  Brian M. Scott Mar 29 '13 at 17:49

A set $U$ is neither open nor closed implicitly. In the real line $[0,1)$ is neither open nor closed, but in the interval $[0,1]$ it is open. Whether a set is considered open is relative to a parent topology.

So we say $U$ is open/closed in another space $X$. Sometimes, the $X$ is implicit, but when somebody tells you a set $U$ is open, there is always either an explicit or implied $X$.

For example, if $U$ is the set of even numbers, you might think "$U$ isn't open." But it is, when considered as a subset of the integers. The even numbers are not open in the set of real numbers, but that is not a contradiction.

The integers, then, are closed as a subset of the real line. As you rightly note, the integers are not closed as a subset of the extended real line, but I doubt Rudin is asserting that, and there is no contradiction there.

There are some general things you can say. If $U\subset V\subset W$ are topological spaces, then if $U$ is open (closed) in $V$ and $V$ is open (resp. closed) in $W$, then $U$ is open (resp. closed) in $W$.

This doesn't help, though, because if $U$ is the integers, $V$ is the real line, and $W$ is the extended real line, we can't conclude anything from $U$ being closed in $V$ because $V$ is not closed in $W$.

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So does a limit point have to exist in the parent topology to be considered a limit point? –  DanZimm Mar 29 '13 at 17:19
    
Yes. When dealing with limit points, we are always talking about in the given topology - there are lots of different ways to "extend" the real line - for example, we can extend the real line with a single $\infty$. There no "natural" extension of the real line that is forced on us. We like the extended real line because it is useful, but inside any topological space $X$, limit points are in $X$. –  Thomas Andrews Mar 29 '13 at 17:22
    
Just realized it does ^ stupid question –  DanZimm Mar 29 '13 at 17:25
    
It's not a stupid question. Rudin's definitions took me some time to get used to. One of the confusing things in topology is that "closed" and "open" are not opposites. If you know a set is not closed, you don't know it is open, and a set can be both closed and open. You have to get used to that - it can be a bit perplexing at first. –  Thomas Andrews Mar 29 '13 at 17:33

A limit point in a metric space $X$ of a subset $A\subset X$ is a point $x\in X$ for which a sequence $(a_i)_{i\geq 0}$ converges to $x$, with $a_i\in A$ for all $i\geq 0$. The only convergent sequences of elements of $\mathbb{Z}$ in $\mathbb{R}$ (that is, with the induced topology/metric) are sequences which are eventually constant, further the limit point of such a sequence is simply the integer at which it is constant. It follows that $\mathbb{Z}$ contains all of its limit points and is closed.

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a limit point is not defined by using a sequence, a sequence is defined using a limit point... how can you then claim this? –  DanZimm Mar 29 '13 at 16:57
    
I should have been more precise. A limit point of a **subset $A$ of a metric space $X$** is a point $x\in X$ for which there exists a sequence $(a_i)_{i\geq 0}$ such that $a_i\in A$ for all $i\geq 0$ and the sequence $(a_i)_{i\geq 0}$ has $x$ as a limit. –  Daniel Rust Mar 29 '13 at 17:03
    
@Dan: A sequence is simply a function whose domain is $\{k:k\ge n\}$ for some fixed integer $n$, usually $0$ or $1$. It is certainly not true that ‘a sequence is defined using a limit point’. Given a sequence, you can ask whether it converges to some point, and given a point, you can ask whether there is a sequence converging to it, but that’s the only connection. –  Brian M. Scott Mar 29 '13 at 17:04
    
The version in your comment still isn’t quite right. If you want $x$ to be a limit point of the set $A$, you need to be sure that there is a $1$-$1$ sequence in $A$ converging to $x$; otherwise you get only that $x\in\operatorname{cl}A$, possibly for the trivial reason that it’s in $A$. –  Brian M. Scott Mar 29 '13 at 17:05
    
The closure of $A$ in $X$ is the set of all limit points of $A$ in $X$ and $A$ is then closed if and only if it is equal to its closure. I don't see the problem here. –  Daniel Rust Mar 29 '13 at 17:12

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