Simple Topology problem (Munkres)

In summary: Also I should point out that there's a small typo in the first line. It should be "If T_a is a family of topologies on X..." instead of "If {Ta} is a family of topologies on X...". Overall, great job summarizing the conversation! In summary, the conversation discussed a problem from Munkres book (2nd edition) about showing that the intersection of a family of topologies on X is also a topology on X. The latter part of the problem was discussed and a counterexample was given to show that the union of a family of topologies may not necessarily be a topology. The conversation also touched on ways to simplify the proof and a small typo was pointed out.
  • #1
jRSC
2
0
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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  • #2
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 [itex]\emptyset[/itex] 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 [itex]\{T_\alpha\}[/itex] of topologies on a set X. Let,
[tex]T = \bigcap T_\alpha[/tex]
For T to be a topology we must have,
1) [itex]\emptyset,X \in T[/itex].
2) If [itex]U_1,U_2 \in T[/itex], then [itex]U_1 \cap U_2 \in T[/itex].
3) If [itex]\{U_i\}[/itex] is a family of subsets of X such that [itex]U_i \in T[/itex], then [itex]U=\cup U_i \in T[/itex].

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

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

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

3. you gave me.

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

2. Let U3[tex]\subseteq[/tex]U1[tex]\cap[/tex]U2. If U1, U2 are in T, then there must be some T1, T2 such that U1 [tex]\in[/tex] T1 and U2 [tex]\in[/tex] T2. However, there is not necessarily a T3 [tex]\in[/tex] {Ta} such that U3 [tex]\in[/tex] T3, so the family [tex]\cup[/tex]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}. [tex]\cup[/tex]Ta = T1 [tex]\cup[/tex] T2 = {o, X, {a, b}, {b, c}} is not a topology because {a,b} [tex]\cap[/tex] {b, c} = {b} [tex]\notin[/tex] [tex]\cup[/tex] 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)?
 
  • #4
jRSC said:
Thanks, I think I got it.

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

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

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

2. Let U3[tex]\subseteq[/tex]U1[tex]\cap[/tex]U2. If U1, U2 are in T, then there must be some T1, T2 such that U1 [tex]\in[/tex] T1 and U2 [tex]\in[/tex] T2. However, there is not necessarily a T3 [tex]\in[/tex] {Ta} such that U3 [tex]\in[/tex] T3, so the family [tex]\cup[/tex]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}. [tex]\cup[/tex]Ta = T1 [tex]\cup[/tex] T2 = {o, X, {a, b}, {b, c}} is not a topology because {a,b} [tex]\cap[/tex] {b, c} = {b} [tex]\notin[/tex] [tex]\cup[/tex] 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 [itex]\emptyset[/itex] and X to your topologies to get:
[tex]T_1 = \{\emptyset,\{a\},X\} \qquad T_2 = \{\emptyset,\{b\},X\} \qquad T_1 \cup T_2 = \{\emptyset,\{a\},\{b\},X\}[/tex]
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.
 
  • #5


Hello, it seems like you are struggling with understanding the concept of topologies and how to prove them. Let me try to provide some clarification and guidance.

First, it is important to understand the definition of a topology. A topology on a set X is a collection of subsets of X that satisfy certain properties, specifically:

1. The empty set and X itself are both in the topology.
2. The union of any number of sets in the topology is also in the topology.
3. The intersection of a finite number of sets in the topology is also in the topology.

Now, let's look at the problem at hand. You are given a family of topologies, {Ta}, on a set X and you are asked to show that the intersection of these topologies, (intersection)Ta, is also a topology on X. To do this, you need to show that the three properties mentioned above hold for this intersection.

First, the empty set and X itself are both in the intersection of the topologies {Ta}, since they are in each of the individual topologies.

Next, let A and B be two sets in the intersection of the topologies. This means that A and B are in every topology in the family {Ta}. Since each of these topologies is a topology on X, we know that the union of A and B is also in each of the topologies, and therefore in the intersection. This satisfies the second property.

Finally, let A and B be two sets in the intersection of the topologies. This means that A and B are in every topology in the family {Ta}. Since each of these topologies is a topology on X, we know that the intersection of A and B is also in each of the topologies, and therefore in the intersection. This satisfies the third property.

Therefore, we have shown that the intersection of the topologies {Ta} satisfies all three properties of a topology on X, and thus is a topology on X.

As for the second part of the problem, you are asked if the union of the topologies, UTa, is necessarily a topology on X. As you have correctly pointed out, this is not necessarily the case. Your counterexample is valid, and it shows that the union of topologies may not satisfy the third property mentioned above.

I hope this helps clarify the problem for you. Remember to always refer back to the definition of topologies and their properties when trying to prove them.
 

1. What is a simple topology?

A simple topology is a mathematical concept that deals with the properties and relationships of sets and their subsets. It is a way of defining the "closeness" or "connectedness" of points in a set, and it is used to study the properties of spaces such as curves and surfaces.

2. What is the Munkres Topology?

The Munkres Topology, also known as the "Topology on Pointed Sets", is a specific type of topology introduced by James Munkres. It is based on the concept of a pointed set, which is a set with a distinguished point that serves as a base point for defining open sets.

3. What is the significance of the Munkres Topology?

The Munkres Topology is significant because it provides a framework for understanding and analyzing the properties of spaces using simple and intuitive concepts. It is also widely used in various branches of mathematics, including algebraic topology, differential topology, and algebraic geometry.

4. What is the "simple topology problem" in Munkres Topology?

The "simple topology problem" is a specific problem in Munkres Topology that deals with the relationship between the topology of a space and its continuous maps. It states that if two spaces have the same topology, then any continuous map between them must be a homeomorphism (a bijective map that preserves the topology).

5. How is the "simple topology problem" solved?

The "simple topology problem" can be solved by first understanding the definition of a continuous map and a homeomorphism in the context of Munkres Topology. Then, using the properties and relationships of these concepts, one can prove that any continuous map between spaces with the same topology is indeed a homeomorphism, thus solving the problem.

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