## Munkres Chapter 2 Section 18

**1.**

**Problem:** Show that the normal formulation of continuity is equivalent to the open set version.

**Proof:** Suppose that are metric spaces and for every and every there exists some such that . Then, given an open set we have that . To see this let then and since is open by hypothesis there exists some open ball such that and thus by assumption of continuity there is some such that and so and thus is an interior point of .

Conversely, suppose that the preimage of an open set is always open and let and be given. Clearly is open and thus is open. So, since there exists some such that and so

**2.**

**Problem:** Suppose that is continuous. If is a limit point of the subset of , is it necessarily true that is a limit point of ?

**Proof:** No. Consider with the suspace topology inherited from with the usual topology. Define

This is clearly continuous since and which are obviously open. But, notice that is a limit point for since given a neighborhood of we must have that there is some which contains it. But, is not a limit point for since that set has no limit points.

**3.**

**Problem:** Let and denote a singlet set in the two topologies and respectively. Let be the identity function. Show that

**a) ** is continuous if and only if is finer than

**b) ** is a homeomorphism if and only if

**Proof:**

**a)** Assume that is continuous then given we have that . Conversely, if is finer than we have that given that

**b)** If is a homeomorphism we see that both it and are continuous and so mimicking the last argument we see that and . Conversely, if then we now that

or equivalently that

which defines the homeomorphic property.

**4.**

**Problem:** Given and show that the maps and given by and are topological embeddings.

**Proof: **Clearly and are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, then by definition of an ordered pair we must have that . Lastly, the inverse function is continuous since is the restriction of the projection to . The same is true for .

**5.**

**Problem: **Show that with the usual subspace topology and .

**Proof:** Define and . These are easily both proven to be homeomorphisms.

**6.**

**Problem: **Find a function which is continuous at precisely one point.

**Proof:** Define

Suppose that is continuous at , then choosing sequences of rational and irrationals numbers respectively both converging to . We see by the limit formulation of metric space continuity that

And so if were to be continuous anywhere it would have to be at . To show that it is in fact continuous at we let be given then choosing we see that from where the conclusion follows since this implies that .

**7.**

**Problem: **

**a) **Suppose that is “continuous from the right”, that is, for each . Show that is continuous when considered as a function from to .

**b) **Can you conjecture what kind of functions are continuous when considered as maps as . As maps from to ?

**Proof:**

**a) ** Note that by the assumption that we know that for every there exists some such that implies that . So, let be open and let . Then, and since is open we see that there is some such that . But, by assumption there exists some such that . But, and thus and thus and so is an interior point for from where it follows that is open and thus is continuous.

**b) ** I’m not too sure, and not too concerned right now. My initial impression is that if is continuous then is open which should be hard to do. Etc.

**8.**

**Problem:** Let be an ordered set in the order topology. Let be continuous.

**a) **Show that the set is closed in

**b) **Let . Show that is continuous.

**Proof: **

**a)** Let then . Suppose first that there is no and consider

This is clearly open in by the continuity of and is contained in it. Now, to show that let then which with simplification gives the important part that and so but since there is no such that this implies that . Similar analysis shows that and since there is no as was mentioned above this implies that . Thus, and thus .

Now, suppose that there is some such that then letting we once again see that is open and . Furthermore, a quick check shows that if that and so and and so and so so that . The conclusion follows

**b)** Let and . As was shown in a) both are closed and thus define

Notice that since are both assumed continuous and that we may conclude by the gluing lemma that is in fact continuous. But, it is fairly easy to see that

**9.**

**Problem:** Let be a collection of subset of ; let . Let and suppose that is continuous for each

**a) **Show that if the collection is finite each set is closed, then is continuous.

**b)** Find an example where the collection is countable and each is closed but is not continuous.

**c)** An indexed family of sets is said to be locally finite if each point of has a neighborhood that intersects only finitely many elements of . Show that if the family is locally finite and each is closed then is continuous.$

**Proof:**

**a) **This follows since if is closed then it is relatively easy to check that but since each is continuous we see that is closed in . But, since each is closed in it follows that each is closed in . Thus, is the finite union of closed sets in , and thus closed.

**b) **Give the subspace topology inherited from with the usual topology and consider with

Clearly each i

**Lemma:** Let be any topological space and be a locally finite collection of subsets of . Then,

**Proof:** The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let since the collection of sets is locally finite there exists some neighborhood of such that it intersects only finitely many, say , elements of the collection. So, suppose that then is a neighborhood of which does not intersect contradicting the assumption it is in the closure of that set.

Now, once again we let be closed and note that and each is closed in and since is closed in we see that is closed in . So, noting that it is evident from the assumption that is locally finite in that so is and thus (for notational convenience) letting the above lemma implies that

From where it follows that the preimage of a closed set under is closed. The conclusion follows.

**10.**

**Problem:** Let and be continuous functions. Let us define a map by the equation . Show that is continuous.

**Proof:** This follows from noting the two projections of are and . But, both of these are continuous since . To see this we note that if and only if which is true if and only if or in other words . Using this we note that the preimage an open set in will be the product of open sets by the continuity of . It clearly follows both projections, and thus the function itself are continuous.

**11.**

**Problem:** Let . We say that is continuous in eahc variable separately if for each , the map is continuous and for each the map is continuous. Show that if is continuous then is continuous in each variable separately.

**Proof: **If is continuous then clearly it is continuous in each variable since if we denote by the mapping we see that where but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.

**12.**

**Problem:** Let be given by

**a) **Show that is continuous in each variable separately.

**b)** Compute .

**c)** Show that is not continuous

**Proof:**

**a) **Clearly both and are continuous for since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at since it is trivial to check that $

**b) **Evidently

**c)** This clearly proves that is not continuous with is not continuous since if is the diangonal we have that

and so in particular

**13.**

**Problem:** Let ; let be continuous and let be Hausdorff. Prove that **if** may be extended to a continuous function , then is uniquely determined by .

**Proof:** I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways

**Way 1(fun way!):**

**Lemma:** Let be any topological space and a Hausdorff space. Suppose that are continuous and define . Then, is closed in

**Proof:** Note that is clearly continuous since and . It is trivial then to check that and since is Hausdorff we have that is closed and the conclusion follows.

From this we note that if agree on such that we have that

From where it follows that and so . So, thinking of as a subspace of we see that and thus clearly is dense in . So, the conclusion readily follows by noting that if are two continuous extensions then by definition .

**Way 2(unfun way): **Let be two extensions of and suppose there is some . Clearly and thus is a limit point of . So, by assumption and so using the Hausdorffness of we may find disjoint neighborhoods of them respectively. Thus, are neighborhoods of in . Thus, is a neighborhood of . But, clearly there can be no otherwise . It follows that is a neighborhood of disjoint from which contradicts the density of in . The conclusion follow.

A comment on your solution to #2: it appears that defining X=(-1,0)\cup(0,1) is unnecessary, since X=(-1,0) would work just as well. I suspect that your example is a holdover from an early draft of your solution when you were likely trying to consider 0 as the limit point.

At any rate, you also have two typos: f^{-1}(\{0\})=(-1,0) as opposed to f^{-1}\{(1\})=(-1,0), and you have an extra \cap in your third to last tex string.

Comment by soiteroo | July 3, 2011 |

Ah, thank you! You are one-hundred percent correct! It was a holdover from a previous draft. I had done all of these far before I posted them up here and was anxious to get them down, they are also smattered with problems from other texts by accident. I will fix that typo then, thanks!

Comment by Alex Youcis | July 8, 2011 |

Also, though it’s essentially what you’re doing I think 9a is even more trivial when you recognize it as induction on the number of pasted-together sets in the pasting lemma.

[I'm not trying to be disagreeable; I'm going through the same project as you and I've been comparing my solutions to yours.]

Comment by soiteroo | July 3, 2011 |

Friend,

Thanks for the input! I see what you’re saying, and I agree that induction would work fine! I think it is more of a stylistic difference than anything.

I don’t find you disagreeable in the slightest, I really enjoy when people point out flaws in my solutions/alternate (better) solutions! Math is a communal subject, and any feedback (good or bad) helps me learn.

I wish you luck in your goal of finishing Munkres. I eventually got restless and moved on to other subjects, but I got fairly far (I just didn’t post them). If you have any specific questions about a problem feel free to e-mail me!

Comment by Alex Youcis | July 8, 2011 |