HW 2 (p 83: 4, 8)

4a.) Since ∅, X ∈ T_{α} for all α, ∅, X ∈ ∩T_{α}

Suppose U_{β} ∈ ∩T_{α} for all β ∈ B. Then U_{β} ∈ T_{α} for all α, β. Since T_{α} is a topology, ∪_{β∈B}U_{β} ∈ T_{α} for all α. Thus ∪_{β∈B}U_{β} ∈ ∩T_{α}

Suppose U_{i} ∈ ∩T_{α} for i = 1, ..., n. Then U_{i} ∈ T_{α} for all α, i = 1, ..., n. Since T_{α} is a topology, ∩_{i=1}^{n}U_{i} ∈ T_{α} for all α. Thus ∩_{i=1}^{n}U_{i} ∈ ∩T_{α}

Let T_{1} = {∅, {a, b}, {a, b, c}} and T_{2} = {∅, {b, c}, {a, b, c}}. Then T_{1} ∪ T_{2} = {∅, {a, b}, {b, c}, {a, b, c}}. {a, b} ∩ {b, c} = {b} ∉ T_{1} ∪ T_{2}. Thus, T_{1} ∪ T_{2} is not a topology.

4b.) Lemma: ∩T_{α} is the unique largest topology contained in all the T_{α}.

∩T_{α} is a topology contained in all the T_{α}. Suppose T is a topology contained in all the T_{α}. Then T ⊂ T_{α} for all α implies T ⊂ ∩T_{α}. Therefore ∩T_{α} is larger than or equal to all other topologies conntained in all the T_{α} and thus ∩T_{α} is the unique largest topology contained in all the T_{α}.

Lemma: ∪T_{α} is a subbasis for the unique smallest topology containing all the T_{α}.

Since X ∈ T_{α}, ∪_{U_{β}∈∪T_{α}}U_{β} = X. Thus, ∪T_{α} is a subbasis.

Let T be the topology generated by the subbasis ∪T_{α}. Suppose that T′ is a topology containing all the T_{α}. Then ∪T_{α} ⊂ T′. If U ∈ T , then U = ∪_{β∈B}(∩_{i=1}^{n}U_{i,β}) where U_{i,β} ∈ ∪T_{α} ⊂ T′. Hence U ∈ T′, and thus T ⊂ T′. Therefore T is smaller than all or equal to other topologies conntaining all the T_{α} and thus T is the unique smallest topology containing all the T_{α}.

4c.) The largest topology contained in T_{1} and T_{2} = T_{1} ∩ T_{2} = {∅, {a}, {a, b, c}}. A subbasis for the largest topology containing T_{1} and T_{2} = T_{1} ∪ T_{2} = {∅, {a}, {a, b}, {b, c}{a, b, c}}. Thus, the largest topology containing T_{1} and T_{2} = {∅, {a}, {b}, {a, b}, {b, c}, {a, b, c}}.

8a.) Let B = {(a, b) | a < b, a and b rational}. Since (a, b) is open for every a < b, a and b rational, B is a collection of open sets of R with the standard topology. Suppose that U is an open set in R and x ∈ U. Since B′ = {(a, b) | a < b, a and b real numbersl} is a basis for the standard topology and U is open, there exists a, b ∈ R, a < b such that x ∈ (a, b) ⊂ U. Since the rationals are dense in R, there exists c, d such that a < c < x < d < b. Thus x ∈ (c, d) ⊂ U. Since (c, d) ∈ B, B is a basis for the standard topology.

8b.) Let T be the topology generated by C. [π, 4) is open in the lower limit topology since it is a basis element, but [π, 4) is not open in T . π ∈ [π, 4). If π ∈ [a, b) where a, b are rational, then a ≤ π < b. Since a is rational and π is irrational, a ≠ π. Thus a < π < b. Hence \frac{a+π}{2} ∈ [a, b), but \frac{a+π}{2} ∉ [π, 4). Thus [a, b) ⊄ [π, 4). Hence there does not exists a basis element, [a, b) in C such that π ∈ [a, b) ⊂ [π, 4). Thus [π, 4) is not open in T .