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Given the lattice $(X,\cup,\cap)$ of all finite and cofinite subsets of $P(\mathbb{N})$ I have to prove, that this is not a complete lattice.

At first I thought I take the union of all subsets of $X$ which have a single, even element. The resulting set would be the set of all even numbers. This set is not in $X$ because it is neither finite nor cofinite.

But then I realized that I can't do that because this set was never a subset of $X$ in the first place.

So, does there even exist a subset of $X$ which has no join or meet in $X$?

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Let $E=\big\{\{2n\}:n\in\Bbb N\big\}$, the set of all singletons of even natural numbers. Clearly $E\subseteq X$, but $E$ has no supremum in $X$: every set of the form $\Bbb N\setminus F$ such that $F$ is a finite set of odd natural numbers is an upper bound for $E$, and none of these upper bounds is minimal.

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    But isn't $E$ just the set of all even natural numbers, which would be neither finite nor cofinite? – 2017-01-04
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    @de_dust: No, $E$ is the set of all sets $\{2n\}$ for $n\in\Bbb N$, not the set of all natural numbers $2n$ for $n\in\Bbb N$. $E$ is not a subset of $\Bbb N$; it’s a subset of $X$. – 2017-01-04
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    I see, so would it be correct if I write: For every $n\in\mathbb{N}$ we get a set $E_n:=\{\{2\},\{4\},\{6\},...,\{2n\}\}$. Also, for every $n\in\mathbb{N}$, $E_n\in X$. The supremum of $E_n$ would be the greatest even natural number. Since such a number does not exist, $E_n$ doesn't have a supremum and therefore $(X,\cup,\cap)$ is not complete. Is this reasoning correct? – 2017-01-04
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    @de_dust: No, that’s not right. Your sets $E_n$ are not elements of $X$: they are subsets of $X$. The sets $\{0\},\{2\},\{4\},\ldots$ are elements of $X$. $E=\big\{\{0\},\{2\},\{4\},\ldots\big\}$ is a subset of $X$, one that has no least upper bound in $X$. – 2017-01-04
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    My mistake, I meant to write $E_n\subseteq X$. Since, for a complete lattice, there must be a supremum for every subset of $X$. Would it then be correct? – 2017-01-04
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    @de_dust: No, because what you want is *one* subset of $X$ that has no supremum in $X$. This amounts to wanting one family, $E$, of finite subsets of $\Bbb N$ whose union is infinite, so that it’s not a subset of any finite subset of $\Bbb N$, but whose union is not cofinite, so that there is no minimal cofinite subset of $\Bbb N$ that contains it. I gave one of the simplest examples of such a family. – 2017-01-04
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    Ok, another attempt: The supremum of $E$ would be the union of the sets $\{0\}, \{2\}, \{4\}, ...$ which is the set of all even natural numbers. Since this set is neither finite nor cofinite, it is not in $X$ and therefore $(X,\subseteq)$ is not a complete lattice. Would that work? – 2017-01-05
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    @de_dust: That’s closer, but the supremum would not actually have to be $\{2n:n\in\Bbb N\}$. If we change $X$ to $Y$, the set of finite subsets of $\Bbb N$ together with the single infinite set $\Bbb N$, $E$ *would* have a supremum in $Y$: it would be $\Bbb N$. Each $\{2n\}\in E$ is a subset of $\Bbb N$, so $\Bbb N$ is certainly an upper bound for $E$. Moreover, it’s the *only* upper bound for $E$ in $Y$, so it must be the supremum of $E$. What is true is that the members of $X$ that are upper bounds for $E$ are the cofinite sets that contain $\{2n:n\in\Bbb N\}$ as a subset. And the reason ... – 2017-01-05
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    ... that none of these is the *least* upper bound for $E$ is that if $C$ is any one of them, there is some odd number $m\in C$, and $C\setminus\{m\}$ is a smaller upper bound for $E$. Thus, $E$ has no smallest upper bound in $X$. – 2017-01-05
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    Thanks for your answers! I'm starting to get it now. I also just realized I made a mistake in my original post: $X$ is the set of all finite and cofinite subsets of $\mathbb{N}$ _not_ of $P(\mathbb{N})$. Does that change a lot? – 2017-01-05
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    @de_dust: No, it changes nothing: I actually misread the question and have been assuming right along that $X$ was the set of finite and cofinite subsets of $\Bbb N$! But now that you’ve pointed out that you actually wrote $P(\Bbb N)$, I understand the source of your confusion. – 2017-01-05