# Simple Topology problem (Munkres)

#### jRSC

Hey guys, I'm reading Munkres book (2nd edition) and am caught on a problem out of Ch. 2. The problem states:

If {Ta} is a family of topologies on X, show that (intersection)Ta is a topology on X. Is UTa a topology on X?

Sorry for crappy notation; I don't know my way around the symbols yet.

For the latter, I say "not necessarily"; for example, if X={a,b,c}, T1={{a}}, and T2={{b}}, then T1UT2={{a},{b}} is not a topology on X because {a,b} is not in the union. If this is wrong, please correct me. However, I am having a very difficult time with these proofs; for the former part of the question, I know the intersection will yield a "coarsest" subset of X, but proving it is a topology is bewildering me.

Thanx for any help.

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#### rasmhop

For the latter, I say "not necessarily"; for example, if X={a,b,c}, T1={{a}}, and T2={{b}}, then T1UT2={{a},{b}} is not a topology on X because {a,b} is not in the union. If this is wrong, please correct me.
This example is not correct, but the idea is the right one. The problem is that T1 and T2 are not topologies since $\emptyset$ and X are not open in either, but they should be open in both. Try to modify T1 and T2 to include these and see if your idea still works.

For the former consider a family $\{T_\alpha\}$ of topologies on a set X. Let,
$$T = \bigcap T_\alpha$$
For T to be a topology we must have,
1) $\emptyset,X \in T$.
2) If $U_1,U_2 \in T$, then $U_1 \cap U_2 \in T$.
3) If $\{U_i\}$ is a family of subsets of X such that $U_i \in T$, then $U=\cup U_i \in T$.

Let me prove 3 for you and see if you can handle 1 and 2 yourself. Since $U_i \in T$ and T is the intersection of $\{T_\alpha\}$ we must have $U_i \in T_\alpha$ for every $\alpha$. Thus $\{U_i\}$ is a family of sets open in $T_\alpha$, but then since $T_\alpha$ is a topology their union must also be in $T_\alpha$ so we get $U \in T_\alpha$ for every $\alpha$. We have now shown that U is in every $T_\alpha$, but then it must be in their intersection so $U \in T$.

#### jRSC

Thanks, I think I got it.

If T=$$\cap$$Ta, then:
1. 0 (empty set) and X $$\in$$T, since 0, X $$\in$$ all Ta

2. If U1, U2$$\in$$ T, then U1, U2$$\in$$ all Ta; their intersection must be in all Ta as well (as is the defn. of a topology), so U1$$\cap$$U2$$\in$$$$\bigcap$$Ta

3. you gave me.

For the next question:
T=$$\cup$$Ta
1. Holds; since 0, X are in each Ta, they are open in the union.

2. Let U3$$\subseteq$$U1$$\cap$$U2. If U1, U2 are in T, then there must be some T1, T2 such that U1 $$\in$$ T1 and U2 $$\in$$ T2. However, there is not necessarily a T3 $$\in$$ {Ta} such that U3 $$\in$$ T3, so the family $$\cup$$Ta is not (necessarily) a topology.

Ex:

Let T1={0, X, {a,b}} and T2={0, X, {b,c}} be topologies on X={a, b, c}. $$\cup$$Ta = T1 $$\cup$$ T2 = {o, X, {a, b}, {b, c}} is not a topology because {a,b} $$\cap$$ {b, c} = {b} $$\notin$$ $$\cup$$ Ta.

Last but not least, to anyone who cares to reply, is there a way to simplify any of the above (or correct me if I am still incorrect)?

#### rasmhop

Thanks, I think I got it.

If T=$$\cap$$Ta, then:
1. 0 (empty set) and X $$\in$$T, since 0, X $$\in$$ all Ta

2. If U1, U2$$\in$$ T, then U1, U2$$\in$$ all Ta; their intersection must be in all Ta as well (as is the defn. of a topology), so U1$$\cap$$U2$$\in$$$$\bigcap$$Ta
That's correct.

For the next question:
T=$$\cup$$Ta
1. Holds; since 0, X are in each Ta, they are open in the union.

2. Let U3$$\subseteq$$U1$$\cap$$U2. If U1, U2 are in T, then there must be some T1, T2 such that U1 $$\in$$ T1 and U2 $$\in$$ T2. However, there is not necessarily a T3 $$\in$$ {Ta} such that U3 $$\in$$ T3, so the family $$\cup$$Ta is not (necessarily) a topology.

Ex:

Let T1={0, X, {a,b}} and T2={0, X, {b,c}} be topologies on X={a, b, c}. $$\cup$$Ta = T1 $$\cup$$ T2 = {o, X, {a, b}, {b, c}} is not a topology because {a,b} $$\cap$$ {b, c} = {b} $$\notin$$ $$\cup$$ Ta.
This is also correct though I'm not sure why you present 1 and 2. The observations made in these are good to make if you want to find a counterexample, but they are not a necessary part of the proof that the statement is false. For that you just need to find a counterexample, and you have.

As for simplification I believe the first part is basically as simple as possible. Your example isn't bad either, but what I actually thought you'd do was simply add $\emptyset$ and X to your topologies to get:
$$T_1 = \{\emptyset,\{a\},X\} \qquad T_2 = \{\emptyset,\{b\},X\} \qquad T_1 \cup T_2 = \{\emptyset,\{a\},\{b\},X\}$$
such that you could re-use your original argument that {a} and {b} are open in the union, but {a,b} isn't. However your example is just as good just thought I'd mention the alternative.

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