p. 83: 3 [adapted from latex HW of Colin McKinney] Show that the collection T_{c} given in Example 4 of §12 is a topology on the set X. T_{c} := {U|X −U is countable or all of X}. First, we must ensure that X and ∅ are elements of T_{c}. For U = ∅, X − ∅ = X. Thus, ∅ ∈ T_{c}. For U = X, X − X = ∅. The null set is finite, and so it is also countable. Hence X ∈ T_{c}. We must also show that T_{c} is closed under arbitrary unions. Take U_{α} ∈ T_{c}. Since X − U_{α} can be either countable or all of X, we must consider two cases. Case 1 : There exists an α_{0} such that X − U_{α_{0}} countable. X − ∪U_{α} = ∩(X − U_{α}), and since X − U_{α_{0}} is countable, and ∩(X − U_{α}) ⊂ X − U_{α_{0}}, it follows that ∩(X − U_{α}) is countable. Hence X − ∪U_{α} is countable, and so in this case ∪U_{α} ∈ T_{c}. Case 2 : X − U_{α} is all of X for all α. Hence U_{α} = ∅ for all α. It follows that ∪U_{α} = ∅ ∈ T_{c} Hence T_{c} is closed under arbitrary unions. Lastly we must show that T_{c} is closed under finite intersections. Take U_{i} ∈ T_{c}. Case 1 : X − U_{i} countable for all i. A finite union of countable sets is also countable. By Problem 1.0 of Homework 1, we may write

X − ⋂_{i=1}^{n} U_{i} = ⋃_{i=1}^{n}(X − U_{i})  (1)

Since ∪(X − U_{i}) is countable, the left hand side of (1) is also countable. Thus, ∪_{i=1}^{n}U_{i} ∈ T_{c} in this case. Case 2 : X − U_{i_{0}} is all of X for some i_{0} ∈ {1, ..., 1}. Hence U_{i_{0}} = ∅, and ∩_{i=1}^{n}U_{i} = ∅ ∈ T_{c}. Thus, T_{c} is closed under finite intersections. T_{c} has thus satisfied all the requirements to be a topology.

Is T_{∞} := {U|X − U is infinite, empty, or all of X} a topology on X? If X is infinite, U_{x} = {x} ∈ T_{∞}, where x is any element of X since X − U_{x} would remain infinite. However, let us define an indexing set Y := X − {x_{1}, x_{2}}, where x_{1} and x_{2} are two distinct elements of X. Hence

X − ⋃_{x∈Y} U_{x} = {x_{1}, x_{2}}.  (2)

Since {x_{1}, x_{2}} is finite, X − ∪_{x∈Y} U_{x} is finite. Hence ∪_{x∈Y} U_{x} is not an element of T_{∞}. Thus when X is infinite, T_{∞} is not a topology since it is not cloased under unions.

Suppose X is finite. Note ∅ and X are elements of T_{∞}. If U = ∅, then X − U = X, so ∅ ∈ T_{∞}. If U = X, then X − X = ∅, and so X ∈ T_{∞}. Let U ∈ T_{∞}. Since X is finite, X − U is not infinite. Thus X − U = ∅ or X − U = X. Thus, U = X or U = ∅. Hence T_{∞} is the indiscrete topology and is thus a topology.