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I would like to clarify the definition of the co-finite topology. The general definition says this:

Let $X$ be a non empty set. Then the collection of subsets of $X$ whose compliments are finite along with the empty set forms a topology on $X$, and is called the co-finite topology.

There is also the example of this statement: $$ \tau = \{\varnothing, \{1\}, \{2\}, \{3\}, \{1,2\}, \{2,3\}, \{1,3\}, X\} $$ which is a co-finite topology because the compliments of all the subsets of X are finite.

Generally, I want to ask one question: why is the complement of each subset of $X$ finite? For example subset which contains only $\{1\}$ for example, complement of this subset is all number except $1$ right? then why is this finite?

Caleb Stanford
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dato datuashvili
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  • Remember that complement is with respect to the full set $X$, which in the example seems to be ${1,2,3}$ (the cofinite topology is the discrete topology on a finite set). – Tobias Kildetoft Aug 23 '13 at 11:28
  • Somewhere it ought to say that $X$ is a finite set; otherwise, most of those sets don't have a finite complement. – Gerry Myerson Aug 23 '13 at 11:28

3 Answers3

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Check, that in order, that $\tau$ is the cofinite topology, the complement of $\{1\}$ has to be finite, so $X$ has to be finite.

On a finite set, the cofinite topology is simply $2^{X}$, the discrete topology, since any subset has a finite complement. So, if $\tau$ is a cofinite topology, then $X=\{1,2,3\}$. If $X$ is supposed to be some other set, then $\tau$ is not the cofinite topology.

Tomas
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  • if $X={1,2,3}$ then complement of $X$ is finite? – dato datuashvili Aug 23 '13 at 11:33
  • The complement of a set $A\subset X$ is $X\setminus A$. So the complement of ${1}$ is ${2,3}$, if $X={1,2,3}$. If $X$ is some infinite set, then $X\setminus{1}$ is cleary infinite, too, but then ${1}$ is not cofinite. You did not tell us, what $X$ is supposed to be. – Tomas Aug 23 '13 at 11:35
  • so it means that if set is finite,then complement of subsets of this set is also member of original set – dato datuashvili Aug 23 '13 at 11:38
  • @dato complements are, as has already been mentioned, always inside some fixed set. In this case, everything is inside the fixed set $X$. – Tobias Kildetoft Aug 23 '13 at 11:41
  • thanks in advance,i am new on topology,started today from this site http://www.emathzone.com/tutorials/general-topology/limit-point-of-a-set.html – dato datuashvili Aug 23 '13 at 11:46
  • @dato Given this question, you might want to take a bit of time to brush up on basic set theory before continuing with topology. It will be a big help along the way. – Tobias Kildetoft Aug 23 '13 at 11:51
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    @dato: Then I assume, you got your question from here, where it clearly says $X={1,2,3}$. You should have either included this into the question or, better, the link. Also, this example for a cofinite topology is not the best one, since, as already mentioned, a finite $X$ is somewhat a degenerate case. Also, this site does not look like the best introduction for topology (no offense), maybe look here for some good introductions. – Tomas Aug 23 '13 at 11:55
  • @dato I agree with Tomas that this does not seem like a good resource to learn from. It is filled with bad grammar and the page about the cofinite topology is not actually correct (it even contradicts itself). – Tobias Kildetoft Aug 23 '13 at 12:01
  • could you give me some good link which is easy for learn – dato datuashvili Aug 23 '13 at 12:05
  • Indeed! Reading the whole article made me shiver. Note, that the cofinite topology contains all cofinite sets, not only some. I provided a link in my last comment above this. – Tomas Aug 23 '13 at 12:06
  • i found it and downloaded,thanks for link – dato datuashvili Aug 23 '13 at 12:18
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The claim is not that the complement of every subset of a set is finite. Rather, the collection of a subsets of a given set whose complement is finite forms a topology. This is not actually hard to check. If a bunch of subsets all have finite complement then certainly their union does so, too. Similarly, for finite intersections.

Rasmus
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  • and do this subsets have finite complements?or complement of set itself is finite – dato datuashvili Aug 23 '13 at 11:29
  • @dato: it is important to notice that one can only talk about the complement of a subset of another given set, see http://en.wikipedia.org/wiki/Complement_%28set_theory%29. – Rasmus Aug 23 '13 at 11:35
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In the above example, the set {1} is open which implies that {1}' is finite,this is only possible when X is finite.......and remember that the complement of a subset of a finite set is always finite,that is why in your example every subset of X is open.

shah faisal
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