Munkres Chapter 2 Section 17
Problem: Prove that every simply ordered set with the order topology is Hausdorff. The product of two Hausdorff spaces is Hausdorff. A subspace of a Hausdorff space is Hausdorff.
Proof: Let have the order topology and let be distinct and assume WLOG that . If there does not exists a such that then are disjoint neighborhoods of respectively. They are disjoint for to suppose that would imply that contradictory to our assumption. If there is some then are disjoint neighborhoods of respectively.
Let be Hausdorff and be distinct. Since are distinct it follows we may assume WLOG that . But, since is Hausdorff there exists disjoint neighborhoods of . So, consider these are clearly neighborhoods of in and to assume would imply .
Lastly, suppose that is Hausdorff and let be a subspace of . If are distinct there exists disjoint neighborhoods of them in . Thus, are disjoint neighborhoods of them in .
Problem: Let be a collection of subsets of the set . Suppose that , and that the finite union and arbitrary intersection of elements of are in . Prove that the collection is a topology on .
Proof: Clearly and are in . Now, suppose that then and since and is closed under arbitrary intersection it follows that and thus . Lastly, suppose that then and since and is closed under finite union it follows that and thus .
Problem: Show that if is a closed in and is closed in that is closed in .
Proof: Since is closed in and is a subspace of we have that for some closed set but since is closed it follows that is the intersection of two closed sets in and thus closed in .
Problem: Show that if is closed in and is closed in then is closed in
Proof: This follows immediately from question 9.
Problem: Show that if is open in and is closed in , then is open in and is closed in .
Proof: This follows immediately from the fact that and .
Problem: Let be an ordered set in the order topology. Show that . Under what conditions does equality hold?
Proof: Let then is a neighborhood of disjoint from and thus . Equality will hold when are limit points of or said otherwise whenever we have that there are some $lated d,e$ such that .
Problem: Let and denote subsets of . Prove that if then , , and . Give an example where this last inclusion is strict.
Proof: We choose to prove the second part first. Let then there is a neighborhood of such that and thus and so . Conversely, suppose that then and and so there are neighborhoods of such that clearly then is a neighborhood of such that and thus .
Using this, if then we have that from where it follows that .
Let then there is a neighborhood of such that and thus for every and thus for every and so finally we may conclude that .
To see when inclusion can be strict consider with the usual topology. Then,
Problem: Criticize proof (see book).
Proof: There is no guarantee that the for which intersected will b e the same that will intersect if you pick another .
Problem: Let and denote subsets of . Determine whether the following equations hold, if any equality fails determine which inclusion holds.
a) Equation does not hold. Consider that but . The inclusion always holds.
b) This follows from a) that equality needn’t hold. Once again the inclusion is true.
c) This need be true either . The inclusion holds.
Problem: Prove that if then in the product topology on .
Proof: Let . Let be a basic open set in which contains . Since we can choose some point . Then, . It follows that
Conversely, let . Let be any set such that . Since is open in it contains some point . Then, . It follows that . A similar technique works for .
10., 11,. 12 Covered in theorem stated and proved at the beginning of the post.
Problem: Prove that is Hausdorff if and only if the diagonal is closed in with the product topology.
Proof: Suppose that is Hausdorff then given we may find disjoint neighborhoods of them. So, and since . It follows that is closed.
Conversely, suppose is closed in and are distinct. Then, and so there exists a basic open neighborhood of such that and so are neighborhoods of in which are disjoint. For, to suppose they were not disjoint would to assume that contradicting the assumption that .
Problem: In the cofinite topology on to what point or points does the sequence converge to?
Proof: It converges to every point of . To see this let be arbitrary and let be any neighborhood of it. Then, is finite and in particular is finite. If it’s empty we’re done, so assume not and let then clearly for all such that we have that and thus is in . The conclusion follows.
Problem: Prove that the axiom is equivalent to the condition that for each pair of points there are neighborhoods of each which doesn’t contain the other.
Proof: Suppose that is then given distinct the sets are obviously neighborhoods of respectively which don’t contain the other.
Conversely, suppose the opposite is true and let then there is a neighborhood of it such that and thus is open and is therefore closed.
Problem: Consider the five topologies on given in exercise 7 of section 13 (my section 2).
a) Deterime the closure of under each of these topologies.
b) Which of these topologies are Hausdorff? Which are
a) As a reminder the topologies are
For the first one we easily see that .
For the second one we can see that . To see this note that is open by definition and thus being the complement of it is closed. Thus,
For the third one . To see this note that we in a sense already proved this in 14, but for any and any neighborhood of it we have that and thus if (assuming it’s non-empty) we see that . Thus, given any point of and any neighborhood of it we have that is infinite, and thus clearly . The conclusion follows from that.
For the fourth topology we note that which is open in with the upper limit topology. Remember that
For the last one . Clearly and since is closed and is the intersection of all closed supersets of it follows that . Now, suppose that then which is a neighborhood of which doesn’t intersect and thus . So, . The conclusion follows.
Lemma: If is Hausdorff, then it is
Proof: Let and by assumption there exists disjoint neighborhoods of respectively and so clearly and thus is open and so is closed.
: Clearly the first topology is Hausdorff since it is an order topology and we proved in the initial post that all order topologies are Hausdorff.
Lemma: Let be a set and two topologies on such that is finer than . Then, if is Hausdorff with the topology it is Hausdorff with the topology.
Proof: Clearly if are distinct we may find disjoint neighborhoods of them in the topology given by and thus the same neighborhoods work in consideration of the topology given by .
: From this lemma it follows that having a finer topology than with the usual topology is Hausdorff
: The cofinite topology is but not Hausdorff. To see that it’s it suffices to prove that is closed for any . But, this is trivial since being the complement of a finite set is open, thus closed. But, it is not Hausdorff since it cannot support two disjoint, non-empty open sets. To see this suppose that is open in with the cofinite topology then is finite and since a disjoint set would have to be a subset of it follows that is finite and thus it’s complement not finite. Thus, is not open.
: Once again this topology finer than that of with the usual topology since
: This isn’t even . To see this we must merely note that if is any set containing we must have that there is some basic open set such that , but this means that . So, there does not exist a neighborhood of which does not contain .
Problem: Consider the lower limit topology on and the topology given by the basis . Determine the closure of the intervals and in these two topologies.
Proof: We first prove more generally that if then . To see this we first note that is open since which we claim is open. To see this we note that . So, since is the intersection of all closed supersets of and is closed we see that . So, we finish the argument by showing that . To see this we show that is a limit point for from where the conclusion will follow. So, let be any neighborhood of , then we may find some basic open neighborhood such that , but clearly contains infinitely many points from where it follows that is a limit point of . From this we may conclude for that and
We now claim that in the topology generated by that and . More generall, let us prove that in this topology
Clearly if then choosing some rational number and some rational number then and so that . Also, if then choosing some such that we see that and . Thus, the only possibilities for are . But, just as before if is any neighborhood we may find some open basic neighborhood $latex [p,q)$ such that but clearly and thus since was arbitrary it follows that .
So, now we split into the two cases. First assume that then given any neighborhood of we may find some basic neighborhood such that , but since we see that from where it follows that and thus since was arbitrary it follows that and thus . Now suppose that , then is clearly a neighborhood of that doesn’t intersect and thus so that
The conclusion follows.
Problem: If , we define the boundary of by .
a) Prove that and are disjoint and
b) Prove that i and only if is both open and closed
c) Prove that is open if and only if
d) If is open, is it true that ?
a) Clearly if then there exists a neighborhood of whose intersection with the complement of is empty, thus . Now, let now if there exists a neighborhood of such that then and if not then every neighborhood of contains points of and since it follows that it also contains points of . Thus, either or , thus . Conversely, if then either there exists a neighborhood of such that and thus . Conversely, if then for every neighborhood of we have that and in particular and so .
b) Suppose first that . If we’re done, so assume not and let . Since there is a neighborhood of such that but since it follows that and thus is open. Conversely, letting we see that and so by the same logic there exists a neighborhood of such that and thus is open and so closed.
Conversely, suppose that is both open and closed and suppose that . If then for every neighborhood of we must have that and thus , which is a contradiction. Conversely, if then for every neighborhood of we must have that and thus which is a contradiction.
c) Suppose that is open and let , then every neighborhood of contains points of by definition, but since (it can’t be an interior point, thus ) they must be points of and thus . So, . Conversely, if then must be a limit point of which is not in and thus every neighborhood contains points of and so
d) No, it is not true. With the usual topology on the set is open, but
We choose to omit 18, 19, 21 on the grounds that they are monotonous and easy, monotonous and easy, and way, way to long.
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