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I didn't want my first question here to be that stupid, but this is something that bothers me very much as it seems simple and yet I get confused when reading and thinking over it. Not clearly understood basics always get me.

Here are some details. I have this problem:

Let $X$ be $\Bbb R$, and let $\Omega$ consist of the empty set and the complements of all finite subsets of $\Bbb R$. Is $\Omega$ a topological structure?

Before that I solved another one:

Let $X$ be $\Bbb R$, and let $\Omega$ consist of the empty set and all infinite subsets of $\Bbb R$. Is $\Omega$ a topological structure?

I proved it by assuming that $[a, b]$ is a finite subset without giving it much thought, so I looked at all of the five other types of intervals.

As I started expanding on this I said, OK, here's the common form of all finite subsets of $\Bbb R$: $[a, b]$, when it struck me: there should be an infinite number of points in this interval (e.g. in $[0, 1]$, the ray 1/n, n -> Infinity), so it must not be finite, which means that I should look only at explicitly defined intervals, e.g., $[a < b < c < d < \cdots]$ where the number of elements is finite. And here I'm lost.

I found this question: Closed Infinite intervals where the different types of infinite intervals were described in the answer and the form $[a, b]$ was not listed.

So common sense tells me the a closed interval is infinite, but I cannot find it written anywhere. (I may have also other faults in my thinking but I'd appreciate if you could help me with the main question.)

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    $\begingroup$ There are infinitely many points in $[0,1]$, so, no, it is not finite. It is bounded, which can feel intuitively like "finite," but that is not what "$[0,1]$ is finite" means. $\endgroup$ – Thomas Andrews Sep 5 '13 at 16:43
  • $\begingroup$ What's this about a ray 1/n, n -> Infinity? What does that mean? $\endgroup$ – dfeuer Sep 5 '13 at 16:52
  • $\begingroup$ What the other post means by "infinite intervals" is "unbounded intervals." They aren't talking about cardinality. $\endgroup$ – Cameron Buie Sep 5 '13 at 16:59
  • $\begingroup$ @dfeuer: I wanted to say a sequence. The sequence of numbers of the form 1/n where n goes to Infinity is contained in the [0, 1] interval for example and that sequence is infinite. I hope this is a bit more clearer, although it still might not be that exact.. the guys in the answers gave more thorough definitions. $\endgroup$ – simich Sep 5 '13 at 17:02
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Note that if $[a,b]$ is a closed interval and $a=b$ then $[a,b]$ consists only of one point, $\{a\}$. However if $a<b$ then we have that $f(n)=a+\frac{b-a}n$ is an injection from $\Bbb N$ into $[a,b]$.

Moreover, if you already know that $(a,b)\subseteq[a,b]$ and that the former is infinite, then the latter must be infinite.

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    $\begingroup$ Yeah, I think his real confusion is that hew views bounded sets as finite. A relatively common mistake amongst people not used to dealing with infinite cardinals. $\endgroup$ – Thomas Andrews Sep 5 '13 at 16:45
  • $\begingroup$ @Thomas: Well, measure and cardinality are not the same. $\endgroup$ – Asaf Karagila Sep 5 '13 at 16:52
  • $\begingroup$ Of course, I'm just commenting on the confusion he is likely having, not that he is correct. $\endgroup$ – Thomas Andrews Sep 5 '13 at 16:53
  • $\begingroup$ (And I said nothing about measure.) $\endgroup$ – Thomas Andrews Sep 5 '13 at 16:59
  • $\begingroup$ @ThomasAndrews: Yes, that was my confusion. I had this intuitive understanding that "a bounded interval is finite" - just that. Which I now know is naive. $\endgroup$ – simich Sep 5 '13 at 17:06
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A closed interval $[a, b]$ is defined as the set of all $x$ with $a \le x \le b$. It's infinite except in the special case that $a=b$, when it contains exactly one point.

It's easy to see that when $a<b$ it must be infinite, because, suppose the contrary, that it's finite. Then we'd have $[a,b] = \{x_1, x_2, \ldots, x_n\}$ for some real numbers $x_1 < x_2 < \cdots < x_n$. But obviously there are a lot of other elements of $[a, b]$ that are not enumerated there; for example, $\frac12(x_1 + x_2)$ lies in between $x_1$ and $x_2$, and we left it out; more generally if $0\lt t\lt 1$ then $tx_1 + (1-t)x_2$ is an element of $[a, b]$ that is missing from the purported enumeration $\{x_1, \ldots, x_n\}$. So we have a contradiction, and in fact the enumeration is incomplete, because $[a,b]$ is actually infinite.

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For this question we are just considering $\Bbb R$ to be a collection of points. Intervals are nothing special in the topology, but can be a useful way to define subsets. For your second consider $[0,1]\cup \{2\}$ and $\{2\}\cup [3,4]$. Both are infinite, but is your topological structure supposed to be closed under finite intersections? To prove the first is a structure, you need to show that it is closed under finite intersections. Since the complement of an open set is finite, and the complement of an intersection is the union of the complements...

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You're right about a closed interval of R: it is infinite. But there are other types of subsets of R than closed intervals: you could just have the subset A = {0,3,4} with 3 elements which is clearly finite.

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  • $\begingroup$ What do you mean by "the complement of all finite subsets of $\Bbb R$"? Also, "the union of all finite subsets of $\Bbb R$" is simply $\Bbb R,$ whose complement (in $\Bbb R$) is certainly finite. Finally, the $\Omega$ described in the previous result is not a topology, but the $\Omega$ described in this result is, so the previous result will be of no use here. $\endgroup$ – Cameron Buie Sep 5 '13 at 17:04
  • $\begingroup$ I mean the union of the complements of all finite subsets, and I did quickly realise what you say about the union of all finite subsets, but then my Dad came in form work, so I couldn't correct my glaring error in time to stop all the internet world seeing it... $\endgroup$ – George Tomlinson Sep 5 '13 at 17:53
  • $\begingroup$ Fortunately for you, I think that most of the internet world is not on this site. ;-) $\endgroup$ – Cameron Buie Sep 6 '13 at 5:40
  • $\begingroup$ lol isn't long enough $\endgroup$ – George Tomlinson Sep 6 '13 at 14:07

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