My attempt to prove the following is below:

Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$

My book calls this set the "finite complement topology"

I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly)

Using the definition of topology:

$\mathscr{T}$ is a topology on X if and only if the following are true:
(i) X and Ø are elements of $\mathscr{T}$.
(ii) $\mathscr{T}$ is closed under finite intersections.
(iii) $\mathscr{T}$ is closed under arbitrary unions.

Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2...$ are all finite sets)
So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.

Now to verify the definition.

(i) X and Ø are elements of $\mathscr{T}$.

Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.

(ii) $\mathscr{T}$ is closed under finite intersections.

$U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| < |\mathbb{N}| $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.

(iii) $\mathscr{T}$ is closed under arbitrary unions.

$U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| < |\mathbb{N}| $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.
Now let's consider the infinite case: $V_k \cap ... \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap ... $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.

  • $\begingroup$ Your proof of infinite unions is clever,but not really necessary-a direct proof via De Morgan's laws is a lot easier. Otherwise, this proof looks fine. $\endgroup$ – Mathemagician1234 Jun 20 '16 at 21:27

(i) is fine, (ii) is fine too, although you can do the finite case in one go (not go via 2), as "De Morgan" holds for any family of sets. The $U_i, V_i$ notation is a bit weird, and you need to consider the special case set as well:

Let $U_1,\ldots U_n$ be open in the topology. We want to show the intersection is open, so if any of them is empty, the intersection will be too, so we can discard those cases: assume all $U_i$ are non-empty and open, so their complements are finite. Now De Morgan says:

$$X \setminus (\cap_{i=1}^n U_i ) = \cup_{i=1}^n (X \setminus U_i)$$

and so the right hand side is finite, as a finite union of finite sets, and so $\cap_{i=1}^n U_i$ is cofinite, and hence open.

Now unions, which is more straightforward than you seem to think (no proof by contradiction is needed):

Suppose $U_i, i \in I$ are open. We want to see that $\cup_i U_i$ is open, so we can discard all $U_i$ that happen to be empty, because they do not affect the union. So assume all $U_i$ are cofinite, and we can assume we have at least one (so $I$ is non-empty, or we have an empty union, which is open, so OK).

Now de Morgan again:

$$X \setminus (\cup_i U_i) = \cap_i (X \setminus U_i)$$

and we see that the right hand set is an intersection of at least one finite set (and the intersection only can become smaller with more sets) so the right hand side is surely finite, and $\cup_i U_i$ is cofinite, and we're done.

  • $\begingroup$ Thanks, i have a question. You conclud each part with the saying the sets are open, and I know topologies are open sets. However, I thought closure of $\mathscr{T}_1$ is what needs to be shown; I thought the conclusion would be all sets created by unions/intersections are still part of the original definition of the topology. $\endgroup$ – Michael Maliszesky Jun 21 '16 at 10:50
  • $\begingroup$ That's what I mean by open here: being in the collection we defined. $\endgroup$ – Henno Brandsma Jun 21 '16 at 15:20

To show that an arbitrary union is in the topology, it is simpler to tackle every case directly:

Let $A=\bigcup_{i \in I} U_i$ be an arbitrary union of sets whose complements are are finite. Then by De Morgan's law, $A=X-(\bigcap_{i \in I}V_i)$, for some finite sets $V_i$. Now by the definition of intersection, $\bigcap_{i \in I}V_i \subseteq V_i$ for all $i \in I$. But since $V_i$ is finite, so is any one of its subsets. Hence the complement of $A$ is finite, implying that $A$ is an open set.

Otherwise your arguments look good.

  • $\begingroup$ Thanks, I'm trying to understand your way of showing it. (A lot of this notation is new to me). When you say $i \in I$ what is $I$? $\endgroup$ – Michael Maliszesky Jun 21 '16 at 3:43
  • $\begingroup$ $I$ is an indexing set. When we say arbitrary union we mean a union over possibly uncountably many sets. Using an arbitrary indexing set allows us to bypass specifying how many sets we are taking the union of $\endgroup$ – leibnewtz Jun 21 '16 at 4:53

The problem is that, the set of finite subsets of $X$ may not be countable. So, saying $V_1, V_2, \dots$ are finite subsets is a little bit problematic.

For $(ii)$, you can simply say that:

Let $U_1, U_2 \in \mathscr T$. Let $U = U_1 \cap U_2$. Then $U^c = U_1^c \cup U_2^c$ is a union of two finite sets. Hence $U^c$ is finite and thus $U \in \mathscr T$.

For $(iii)$,

$Let \{U_\alpha \}$ be a set of opens in $\mathscr T$ (with $U_\alpha \neq \emptyset$). Let $U = \bigcup U_\alpha$. Then, $U^c = \bigcap U_\alpha^c$ is an intersection of finite sets, which is also finite. Hence $U^c$ is finite. Therefore $U \in \mathscr T$.

  • $\begingroup$ I don't see how the cardinality affects the OP's proof since he/she's dealing with the intersections of this family of sets, which is always a subset of each of the sets, So if the intersection is infinite, so must each of the individual sets and this would be true regardless of the number of sets involved. In this particular case, because of how the OP's proof proceeds, it doesn't matter. $\endgroup$ – Mathemagician1234 Jun 20 '16 at 21:29

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