Introduction to Topology and Modern Analysis Problems Thread

**Poophead** · April 22nd 2010, 07:06 PM

I've posted a couple of threads asking for help with the exercises in ITMA in the past month, so rather than continue to clog up the forum with new threads every time I need help I figured I'd just make this thread.

I will not only ask for help in this thread, but will also give assistance to anyone who needs help with a part of the book that I've already worked through. Admittedly, this thread is mostly for my benefit since I'm still on the first chapter, but as I progress I will become more and more able to help others who are trying to work through the book.

So I'm almost finished with the first chapter, but am stuck on the very last problem (Section 8, Problem 10). I feel that I've made it almost to the finish line only to collapse. I've already shown that any uncountable set can be represented as the union of a disjoint class of countably infinite sets, and yet I can't show that U(X_i) has cardinality less than or equal to that of I. If someone could give me the nudge that I need to finish this thing off, I'd greatly appreciate it.

Thanks,

Poophead

**Drexel28** · April 22nd 2010, 08:56 PM

Originally Posted by Poophead

I've posted a couple of threads asking for help with the exercises in ITMA in the past month, so rather than continue to clog up the forum with new threads every time I need help I figured I'd just make this thread.

I will not only ask for help in this thread, but will also give assistance to anyone who needs help with a part of the book that I've already worked through. Admittedly, this thread is mostly for my benefit since I'm still on the first chapter, but as I progress I will become more and more able to help others who are trying to work through the book.

So I'm almost finished with the first chapter, but am stuck on the very last problem (Section 8, Problem 10). I feel that I've made it almost to the finish line only to collapse. I've already shown that any uncountable set can be represented as the union of a disjoint class of countably infinite sets, and yet I can't show that U(X_i) has cardinality less than or equal to that of I. If someone could give me the nudge that I need to finish this thing off, I'd greatly appreciate it.

Thanks,

Poophead

You're going to be told to make another thread.

Also, why did you make me go get my book? haha

Question:

Let $\left\{X_i\right\}_{i\in I}$ be any class of countable sets. Show that the cardinal number of $\bigcup_{i\in I}X_i$ is less than or equal to the cardinal number of $I$

The following reflects my own personal taste and might be unappealing to you.

So, if $I\simeq\mathbb{N}$ this is trivial. So, assume not. Then, as you've shown $I=\coprod_{j\in\mathcal{J}}\Omega_j$ where each $\Omega_j\simeq\mathbb{N}$ . So, notice that $\bigcup_{i\in I}X_i=\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega _j}X_i$ but each $\bigcup_{i\in\Omega_j}X_i$ is the countable union of countable sets

and thus countable. So, for each $j\in\mathcal{J}$ there exists some surjection $f_j:\Omega_j\to\bigcup_{i\in\Omega_j}X_i$ and so define $F:\coprod_{j\in\mathcal{J}}\Omega_j\to\bigcup_{j\i n\mathcal{J}}\bigcup_{i\in\Omega_j}X_i:x\mapsto f_j(x)$ where $j$ is the unique element of $\mathcal{J}$ such that $x\in\Omega_j$ .

To see this is a surjection we merely note that if $y\in\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega_j }X_i$ then $y\in\bigcup_{i\in\Omega_{j_0}}X_i$ for some $j_0\in\mathcal{J}$ and so there exists some $x\in\Omega_{j_0}$ such that $x\overset{f_{j_0}}{\longmapsto}y$ . The conclusion follows.

Note the absolute necessity in the above that $I=\coprod_{j\in\mathcal{J}}\Omega_j$ and not $I=\bigcup_{j\in\mathcal{J}}\Omega_j$

**Poophead** · April 22nd 2010, 09:27 PM

Hmmm. Do you think that you could point me to a good reference on basic set theory notation? I'm afraid that I'm not familiar with all of the notation that you've used. This is probably because I decided to dive into this book after teaching myself basic differential and integral calculus.

**Drexel28** · April 22nd 2010, 09:33 PM

Originally Posted by Poophead

Hmmm. Do you think that you could point me to a good reference on basic set theory notation? I'm afraid that I'm not familiar with all of the notation that you've used. This is probably because I decided to dive into this book after teaching myself basic differential and integral calculus.

Which isn't familiar? If it's $\coprod$ then that's my fault for being lazy. It has a technical meaning which is irrelevant. In this context I merely meant that $I=\bigcup_{j\in\mathcal{J}}\Omega_j$ and if $j_1\ne j_2$ then $\Omega_{j_1}\cap\Omega_{j_2}=\varnothing$ which is what you proved, that it can be presented as the union of a disjoint class of sets. That symbol is the "disjoint union" or "coproduct"

Oh, and $\simeq$ just means that they have the same cardinality.

**Poophead** · April 22nd 2010, 09:43 PM

Ah, yes, that was the bit that confused me. Now I understand all of your notation (I had to look up surjection, but I see that a surjection is just a mapping which is onto but not necessarily one-to-one).

I've read through your proof a couple of times and my head is in the process of wrapping around it, so I will be attempting to regurgitate it to you shortly to make sure that I understand it.

**Drexel28** · April 22nd 2010, 09:45 PM

Originally Posted by Poophead

Ah, yes, that was the bit that confused me. Now I understand all of your notation (I had to look up surjection, but I see that a surjection is just a mapping which is onto but not necessarily one-to-one).

I've read through your proof a couple of times and my head is in the process of wrapping around it, so I will be attempting to regurgitate it to you shortly to make sure that I understand it.

**Poophead** · April 22nd 2010, 10:33 PM

Originally Posted by Drexel28

You're going to be told to make another thread.

Also, why did you make me go get my book? haha

Question:

The following reflects my own personal taste and might be unappealing to you.

So, if $I\simeq\mathbb{N}$ this is trivial. So, assume not. Then, as you've shown $I=\coprod_{j\in\mathcal{J}}\Omega_j$ where each $\Omega_j\simeq\mathbb{N}$ . So, notice that $\bigcup_{i\in I}X_i=\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega _j}X_i$ but each $\bigcup_{i\in\Omega_j}X_i$ is the countable union of countable sets

and thus countable. So, for each $j\in\mathcal{J}$ there exists some surjection $f_j:\Omega_j\to\bigcup_{i\in\Omega_j}X_i$ and so define $F:\coprod_{j\in\mathcal{J}}\Omega_j\to\bigcup_{j\i n\mathcal{J}}\bigcup_{i\in\Omega_j}X_i:x\mapsto f_j(x)$ where $j$ is the unique element of $\mathcal{J}$ such that $x\in\Omega_j$ .

To see this is a surjection we merely note that if $y\in\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega_j }X_i$ then $y\in\bigcup_{i\in\Omega_{j_0}}X_i$ for some $j_0\in\mathcal{J}$ and so there exists some $x\in\Omega_{j_0}$ such that $x\overset{f_{j_0}}{\longmapsto}y$ . The conclusion follows.

Note the absolute necessity in the above that $I=\coprod_{j\in\mathcal{J}}\Omega_j$ and not $I=\bigcup_{j\in\mathcal{J}}\Omega_j$

(Edit: the following is long and awkward and probably reads like watching a toddler try to walk.)

Okay, so I'm sure that you will find this tedious, but I'm going to go through this proof verbally. I tend to think either verbally or visually; I'm not yet adept at thinking in mathematical symbols. My train of thought is usually like a narrated film involving amorphous blobs, points, etc., and since I can't show you this film, I'll just give you the narration. I'm going to start by showing that we can actually represent I as the union of a disjoint class of countable sets. (This part took me a while because I thought that it had to do with the fact that I is an index set.)

If we have any uncountable set U, we can map N into it and this will create a partition of our set into a countably infinite subset A and (U - A), which will be uncountable. Now, we can do the same to (U - A), and we will have a partition of U into two countably infinite subsets and one uncountable subset, etc. So let P be the class of partitions of U constructed in this way. P is partially ordered by the number of partitions that U has been broken into for each p in P. Now a chain in this partially ordered set will have the upper bound 2^U, that is, the power set of U, so we know that P has a maximal element. Now observe that in this maximal element none of the partitions of U can be uncountable, or else we could continue the partitioning process by partitioning the uncountable partition. Therefore the partitioning process eventually results in a partition of U into countably infinite subsets.

So partition I this way. Notice that U_i(X_i) is the union of sets that are indexed by some i in some partition of I. But each i indexes a countable set, and each partition of I contains countably many indexes, so the union of sets of {X_i} given by the indexes in any given partition of I is countable. So, by sleight of hand (

), we can map any partition of I onto the entire union of the sets of {X_i} whose indexes are in that partition, since both the partition and the union are countably infinite and so have the same cardinality.

So we can create a mapping from the partitioned I onto U_i(X_i) by taking each index and associating it with the proper surjection (that is, with the aforementioned surjection which maps the indexes in a partition onto the union of {X_i} whose sets each have an index in that partition), which is unique since each index is in only one partition of I. This is a surjection because, for any element x in one of the sets of {X_i}, it is in a set whose index is in a particular partition of I, and since each partition of I maps onto a subset of U_i(X_i), there is some i in I that will map to that x under the surjection for its partition.

I'm not exactly sure why I had to be a disjoint class of countably infinite sets rather than an uncountable union of countably infinite sets.

**Drexel28** · April 22nd 2010, 11:02 PM

Originally Posted by Poophead

So partition I this way. Notice that U_i(X_i) is the union of sets that are indexed by some i in some partition of I. But each i indexes a countable set, and each partition of I contains countably many indexes, so the union of sets of {X_i} given by the indexes in any given partition of I is countable.

I think what you're saying is that because $I=\coprod_{j\in\mathcal{J}}\Omega_j$ that $\bigcup_{i\in I}X_i=\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega _j}X_i$

You can think about it like this. Imagine you have a cylinder. But, if you look at the base of the cylinder (the circle $C$ ) attached to each point is a "fiber", or a strand (a straight line in this case) which goes from that point at the base to a point at the "ceiling" of the cylinder. So, we can think about each point $c$ in our base circle this fiber $f_c$ attached to it. And, so to get the entirety of the cylinder we need to look at all the fibers in the circle, in other words $\text{Cylinder}=\bigcup_{c\in C}f_c$

It is similar (but not really) to this

So we have that $\text{Cylinder}=\bigcup_{c\in C}f_c$ . But, what if we cut our circle into four non-overlapping portions? Like, think about the circle being centered at the origin and cut it into the four portions based on which part of the circle lies in the four quadrants. What we get is four smaller regions $R_1,R_2,R_3,R_4$ which are disjoint and $R_1\cup R_2\cup R_3\cup R_4=\bigcup_{i=1}^{4}R_i=C$ . So, now with our picture to get the whole cylinder we first look at the fibers above the first region $R_1$ and then add the ones above $R_2$ and so on. So, $\bigcup_{i=1}^{4}\bigcup_{c\in R_i}f_c=\text{Cylinder}=\bigcup_{c\in C}f_c$ .

And that's what gives us the inuition that $\bigcup_{j\in\mathcal{J}}\bigcup_{i\in\Omega_j}X_i =\bigcup_{i\in I}x_i$

So, by sleight of hand (

), we can map any partition of I onto the entire union of the sets of {X_i} whose indexes are in that partition, since both the partition and the union are countably infinite and so have the same cardinality.

So, yes. We have that for each fixed $j\in\mathcal{J}$ that (by what you showed) $\Omega_j$ is countable and since we assumed each $X_i$ is countable it follows that $\bigcup_{i\in\Omega_j}X_i$ is the countable union of countable sets and thus countable. (this is visually just looking at the fibers above one of the regions $\Omega_j$ )

So we can create a mapping from the partitioned I onto U_i(X_i) by taking each index and associating it with the proper surjection (that is, with the aforementioned surjection which maps the indexes in a partition onto the union of {X_i} whose sets each have an index in that partition), which is unique since each index is in only one partition of I. This is a surjection because, for any element x in one of the sets of {X_i}, it is in a set whose index is in a particular partition of I, and since each partition of I maps onto a subset of U_i(X_i), there is some i in I that will map to that x under the surjection for its partition.

Yeah, so the intuition now is that, using the cylinder picture, we've found a way to show that the set of all points in the fibers above each region has a smaller cardinal number than the region below it. And the way we've done this by finding a function such that $f:\text{Region }j\to\text{Set of fibers above region }j$ and from here it's obvious what to do. You give me any point in the cylinder it has to be above one of the regions (on one of the fibers) and so the mapping for that region must hit it (since it's a surjection). So, if we define a mapping $F:\text{All the regions}\to\text{The whole Cylinder (all fibers)}$ by just saying "a point in the whole big region must be in one of the smaller regions, so map it to the point that the map defined on that smaller region would send it to". And, since each of the smaller regions hits all the points it's clear that this mapping hits all the points. Thus, we've defined a map from our region (the indexing set) to the whole cylinder (the union) which hits every point (is a surjection) and so from your previous exercise you may conclude that the cardinal number of the cylinder (union) is less than or equal to that of the region (the indexing set)

I'm not exactly sure why I had to be a disjoint class of countably infinite sets rather than an uncountable union of countably infinite sets.

It came into the part where we talked about splitting up the union.

But, more importantly think about what would have happened when we tried to define our mapping $F$ above. If you had given me a splitting up of the region where the things overlap then to each point it might lie in more than one region and so there would be more than one mapping (one for each region) and so there would be no canonical way to map it (you'd have to pick a function to use and then you get into issues of no longer having surjectivity)

Hope that helped.

**Poophead** · April 22nd 2010, 11:24 PM

Wow, what a great post! Thank you so much.

I think I grok it now. Our class of countable sets is like the points in the cylinder, "indexed" by the points in its circular base.

I see why we have problems if the regions of the cylinder overlap, but since we only have to show a surjection, not necessarily a bijection, to demonstrate that one set has greater cardinality than another, it seems like there ought to be a way to look at the union of regions in the base and just pick some f_c for each point that has more than one f_c to choose from. Is the problem that one of the points in the cylinder might get "left out" because the point on the base that was "supposed" to map to it used the wrong f_c and became "redundant"? (I hope you get what I mean.)

**Drexel28** · April 22nd 2010, 11:26 PM

Originally Posted by Poophead

Wow, what a great post! Thank you so much.

I think I grok it now. Our class of countable sets is like the points in the cylinder, "indexed" by the points in its circular base.

I see why we have problems if the regions of the cylinder overlap, but since we only have to show a surjection, not necessarily a bijection, to demonstrate that one set has greater cardinality than another, it seems like there ought to be a way to look at the union of regions in the base and just pick some f_c for each point that has more than one f_c to choose from. Is the problem that one of the points in the cylinder might get "left out" because the point on the base that was "supposed" to map to it used the wrong f_c and became "redundant"? (I hope you get what I mean.)

Yes! Like I said there are problems with surjectivity! Suppose that $x$ was in two regions and by choosing one over the other a point in our cylinder wasn't mapped to (because $x$ was the only point that mapped to it)

**Poophead** · April 22nd 2010, 11:33 PM

That was a truly amazing explanation. You should consider becoming a professor or teacher or author at some point.

I'm so proud of myself for making it through the first chapter of ITMA. I put hours and hours into those exercises!

Of course, you helped a lot! So thank you.

I'm thinking about going through a different calculus book and going further in it than I did my Stewart textbook while I continue to work through ITMA, just to give myself some more intuition and etc.

Any recommendations? I thought that Integrated Physics and Calculus, Vol. I by Andrew Rex and Martin Jackson looked good. Seems like it would be a nice counter-balance to the very theoretical ITMA. (I do prefer the rigorous theoretical stuff, though, which is why I jumped into ITMA in the first place. But a physics/calc. book might be a little more visual and "pleasant".)

**Drexel28** · April 22nd 2010, 11:40 PM

Originally Posted by Poophead

That was a truly amazing explanation. You should consider becoming a professor or teacher or author at some point.

I have my dad to thank for that.

I'm so proud of myself for making it through the first chapter of ITMA. I put hours and hours into those exercises!

I'm proud of you too! haha The first chapter is brutal in terms of sheer volume of work. It went something like this (for my solutions), mind you I TeXed them but iwas like :35 pages, 25 pages, 20 pages, 15 pages, 5 pages, 2 pages, 3 pages.

Keep with the book, it really is good. It's not the best topology text book if you want to be a hard-core topologist (try Munkres, Bourbaki, or Willard for that) but it really gives a solid grounding in sets and functions and metric spaces

Of course, you helped a lot! So thank you.

Of course, I'm glad to help.

I'm thinking about going through a different calculus book and going further in it than I did my Stewart textbook while I continue to work through ITMA, just to give myself some more intuition and etc.

The intuition I have did not come from calculus text books, but from topology. If you are looking for a good book for visual stuff or just to supplement your work here I'd reccomend:

Spivak (either his calculus book or his calculus on manifolds book...depending on how hard you want)

or A Vector Space Approach to Geometry-Hausner

But, really what you might want is a good analysis book or algebra book. For that I reccoment Rudin (Principles of Mathematical Analysis) or Wade (An Introduction to Analysis) and Abstract Algebra (Dummit and Foote) or Topics in Algebra (Herstein) respectively.

**Poophead** · April 28th 2010, 06:19 PM

Do you have an opinion of Introduction to Analysis by Rosenlicht? It's well-reviewed and very inexpensive, but it's not one of the major analysis texts.

Another book that looked interesting is Introduction to Calculus and Analysis, a two-volume work by Courant.

**Drexel28** · April 28th 2010, 06:50 PM

Ask new questions in new threads. If you ask new questions in the same thread the moderators will get angry.

Originally Posted by Poophead

Do you have an opinion of Introduction to Analysis by Rosenlicht? It's well-reviewed and very inexpensive, but it's not one of the major analysis texts.

Another book that looked interesting is Introduction to Calculus and Analysis, a two-volume work by Courant.

I can't speak to Courant. I hear the name incessantly but I have never read a single book by him/her.

I do own Rosenlicht (the dover version) though and am holding it in my hands. The books I suggested earlier are much more comprehensive and better. In fact, I always found Rosenlicht to be a bit unappealing, but I don't know why. That said, it's not a bad book, and for ten to twelve dollars it's worth the money.

Don't have any new Simmons problems, huh?

**Poophead** · April 29th 2010, 05:57 AM

I'm still going through ITMA, but I have several other things on my plate right now, so I'm still on 2.1. I'm on exercise 4 right now. I may decide to devote more time to ITMA next week, but it will probably take me a month or longer to finish chapter 2. Don't worry--if I run into a problem that I can't solve, I'll post about it.

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