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EDIT: This has been edited so please only take a look at the 2nd question D.

I'm not very good with topology but I would like some advice on my homework and also if possible to verify if what I've done is correct. While I'm still very new to topology I would like some advice to help me figure out the correct answer to those I've gotten incorrect.

The first question. We let $X=\{1,2,3\}$ and a function $N$ on $X$ given by: $N(1)=\{\{1,2\},\{1,2,3\}\}$

$N(2)=\{\{2,3\},\{1,2,3\}\}$ and $N(3)=\{\{2,3\},\{1,2,3\}\}$

A) Prove that $N$ does not define a neighborhood topology on $X.$

My answer: Following from the 4th neighborhood axiom, $\{1\}$ is not contained in $\{2,3\}$ therefore $N$ does not define a neighborhood topology on $X.$

B) Add a minimal number of neighborhoods to the above definition $N$ to indeed obtain a neighborhood topology.

My answer: I make the following changes to obtain a neighborhood topology:

$N(1)=\{\{1,2\},\{1,3\},\{1,2,3\}\},\ N(2)=\{\{1,2\},\{2,3\},\{1,2,3\}\},\\ N(3)=\{\{1,3\},\{2,3\},\{1,2,3\}$

The second question:

Consider the upper limit topology:

$N(x)=\{N \subseteq \Bbb R\mid \exists a,b\in\Bbb R: x \in (a,b] \subseteq N\}$

A) Prove that $N$ is a neighborhood topology on $\Bbb R$ (real numbers)

My answer: I assume this is just where I show the 4 definitions that define a Neighborhood and rewrite it to fit this topology.

B) What does a convergence of a sequence in this topology mean?

My answer: If $X_n$ converges to $x$ in this topology then it means:

$ \forall a,b \in \mathbb{R}: a<x \le b: \exists n_0 \in \mathbb{N}: \forall n>n_0: X_n \in (a,b]$

Which basically means that if $X_n$ converges in this topology then it converges in the ordinary topology.

C) Prove or disprove that $1/n$ converges to $0$ in this topology.

My answer: It does not converge because if we have $(-1,0],$ it contains $0$ but it does not contain any points of $1/n.$

D) If ${a_n}$ and ${b_n}$ converge, which of the following limit rules are still valid?

1: $\lim_{n \to \infty}(a_n+b_n)=\lim_{n \to \infty}(a_n)+\lim_{n \to \infty}(b_n)$

2: $\lim_{n \to \infty}(ca_n)=c \lim_{n \to \infty}(a_n)$

3: $\lim_{n \to \infty}(a_n b_n)=\lim_{n \to \infty}(a_n) \lim_{n \to \infty}(b_n)$

4: $\lim_{n \to \infty} \frac{a_n}{b_n}= \frac{\lim_{n \to \infty}(a_n)}{\lim_{n \to \infty}(b_n)}$

5: The existence of $\overline{n}$ with $a_n \le b_n$ for all $n \ge \overline{n}$ implies $\lim_{n \to \infty}(a_n) \le \lim_{n \to \infty}(b_n)$

My answer: I apply 3 "lemmas" which states the following:

1: If $x_n$ converges to x in the upper limit topology then it converges in the ordinary topology.

2: $|x_n > x|=\infty$ implies that $x_n$ does not converge in the upper limit topology

3: $|x_n > x|<\infty$ and $x_n$ converges in the ordinary topology then it also converges in upper limit topology.

This means the following for each rule:

1: This one is still valid and it follows from the 3rd lemma.

2: This one is also valid following from the 3rd lemma however only for $c>0$

3: I don't think this one is valid following the 2nd lemma unless both are greater than 0.

4: I can rewrite $\frac{1}{b_n}=c$. If 3 is false or true then the same applies to this one.

5: I honestly have no idea how to do this so I would like a hint.

Any help or verification is appreciated.

Stefan Hamcke
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Nikolaj Kyed
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  • The intersections of two neighborhood should also be a neighborhood. But it would also suffice to simply add ${1}$ to the neighborhood filter of $1.$ – Stefan Hamcke Sep 21 '13 at 21:48
  • A suggestion: Sometimes the text is easier to read if you use latex-code to type math font. For a basic tutorial see http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Stefan Hamcke Sep 21 '13 at 21:51
  • Hello Stefan, I assume you mean that for B in the first question, so does that mean I should just add {1} to N(2) and N(3)? I also appologise for not using latex code I wasn't sure if it was present here. – Nikolaj Kyed Sep 21 '13 at 21:58
  • No, you just have to add ${1}$ to N$(1)$. Oh, and all supersets of ${1}$, of course. So also ${1,3}.$ – Stefan Hamcke Sep 21 '13 at 21:59
  • Oh okay thank you. If you don't want to look through the 2nd question in it's entirety could you at least look at D and try to give me a clue? If you can't then it's alright. – Nikolaj Kyed Sep 21 '13 at 22:01
  • For B) you should change it to $a<x\le b$. – Stefan Hamcke Sep 21 '13 at 22:04
  • Yes I noticed that I forgot to reverse it from lower to upper limit topology. Now you actually confused me I have to add {1} and {1,3} to N(1)? – Nikolaj Kyed Sep 21 '13 at 22:06
  • Yes, you have to add these two sets to $N(1)$. Another approach is to add ${2}$ and ${1,2}$ to $N(2).$ The point is that ${1,2}$ must contain a smaller nbh of $1$ such that ${1,2}$ is a nbh for every point in that smaller nbh. – Stefan Hamcke Sep 21 '13 at 22:09
  • Oh yea I see it now. For the second question I might have a proof for D, could you look at C and decide if my argument is good enough as proof for 1/n to not converge? – Nikolaj Kyed Sep 21 '13 at 22:11
  • you are right $\frac1n$ does not converge by exactly the reason you describe – Stefan Hamcke Sep 21 '13 at 22:12
  • Thanks you've been a great help. When I'm done with my proof for D, would you be willing to take a look at it? Probably won't have it ready for about 5-10 mins though. – Nikolaj Kyed Sep 21 '13 at 22:14
  • I can take a look at it. Please add the string @StefanH in your comment. That ensures that the person you are replying to is notified by the system. (sometimes it works without that string, but sometimes not) – Stefan Hamcke Sep 21 '13 at 22:16
  • @StefanH. I can see I made a mistake. In D there's supposed be limit with n->infinity (An+Bn)=lim(An)+lim(Bn) – Nikolaj Kyed Sep 21 '13 at 22:25
  • Take $\varepsilon>0$. If the limits $a$ and $b$ exist, then you have to proof that there is an $n_0$ s.t. for all $n>n_0$ we have $a+b-\varepsilon<a_n+b_n\le a+b$. Hint: Apply convergence of $(a_n)_n\to a$ to the value $\varepsilon/2.$ Similiarly, for $(b_n).$ – Stefan Hamcke Sep 21 '13 at 22:36
  • @StefanH. Wouldn't that only make sense for the ordinary topology? – Nikolaj Kyed Sep 21 '13 at 22:38
  • I don't think so. Note that the sets $(a-\varepsilon,\ a]$ form a neighborhood base for $a.$ – Stefan Hamcke Sep 21 '13 at 22:39
  • @StefanH. That strikes me as odd because our TA suggested that we use something he called lemmas (denoted by L). They would state the following: L1: If it converges in upper limit topology then it converges in ordinary topology. L2:|{Xn > x|=inf => Xn does not converge in upper limit topology. L3:|{Xn > x| < inf and Xn converges in ordinary topology then => Xn converges in upper limit topology.

    By then applying the 3rd lemma it should be possible to see that sum of convergence is still converging in upper limit topology.

     – Nikolaj Kyed Sep 21 '13 at 22:42
  • Well, you can use those lemmas, together with the fact that if $(a_n)_n\to a$ and $(b_n)_n\to b$ then $(a_n+b_n)_n\to a+b$ in the ordinary topology. – Stefan Hamcke Sep 21 '13 at 22:58
  • What I gave as hint was similar to the proof as it is done in the ordinary topology. That thing with $\varepsilon/2$ and so ... – Stefan Hamcke Sep 21 '13 at 23:01
  • @StefanH. Thanks, I just think our TA did it to make it a little easier for us to figure out. I still have to prove 4 others, would it be better if I worked with those tonight and then posted it as a separate question tommorrow? – Nikolaj Kyed Sep 21 '13 at 23:04
  • Do you mean four other rules about the sequences or four completely new problems? – Stefan Hamcke Sep 21 '13 at 23:07
  • @StefanH. I mean 4 different limit rules. Multiplication with a constant, product of two convergin sequences, division by two converging sequences and then the existence of a value n with $a_n \le b_n$ for all $N>n$ implying that $\lim{n \to \infty} a_n$ $\le$ $\lim{n \to \infty} b_n$ – Nikolaj Kyed Sep 21 '13 at 23:12
  • I recommend that you add them to your question, as well as your thoughts about them. You can edit your own question any time. Now that you know how to do latex, you can also "latexify" your existing text :-) You can do $\lim_{n\to\infty}$ by typing \lim_{n\to\infty} – Stefan Hamcke Sep 21 '13 at 23:15
  • @StefanH. Does that mean if I edit it, it will show up again as a new question or? – Nikolaj Kyed Sep 21 '13 at 23:16
  • No, it will not show up as a new question. It will only bump the question to the top of the list when you are on the 'active' tab in the question view. – Stefan Hamcke Sep 21 '13 at 23:18
  • @StefanH. Okay thanks, I'll do that tommorrow once I'm done here. Have a good night or day, depending on where you are from :) – Nikolaj Kyed Sep 21 '13 at 23:20
  • That would be good night where I am :-) Thank you, the same to you! – Stefan Hamcke Sep 21 '13 at 23:23

1 Answers1

1

We know that if $a_n\to a,\ b_n\to b$, then $a_n b_n$ converges to $ab$ and $a_n+b_n$ converges to $a+b$ in the ordinary topology $\tau$ on $\Bbb R.$ Also $\frac{a_n}{b_n}$ converges to $\frac ab$ if $b\ne0.$

If $a_n\to a,\ b_n\to b$ in $\lambda$, then they converge in $\tau$, and $a_n, b_n$ are eventually less equal than $a,b$, respectively. Then $a_n b_n\to ab$ in $\tau$, but we can only be sure that $a_nb_n\le ab$ if $a$ and $b$ are both larger than zero. Then it follows that $a_n b_n\to ab$ in the upper limit topology $\lambda.$

And $ca_n\to ca$ also works if $c=0.$

If $a_n\to a,\ b_n\to b\ne0$ in $\lambda$, then they do so in $\tau$, so we have $\frac{a_n}{b_n}\to\frac ab$ in $\tau$. Also $a_n\le a,\ b_n\le b$, and this means that $\frac1b\le\frac1{b_n}$. Now imagine $a<0,\ b>0.$ If $\frac1b\le\frac1{b_n}$, then we can deduce that $a_n\frac1{b_n}\le a\frac 1b.$

Stefan Hamcke
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  • Note that when I wrote $a_n\le a$ I always meant that only finitely many $a_n$ are $>a$, one then says almost all $a_n$ are $\le a$, or $a_n$ is eventually $\le a$. – Stefan Hamcke Sep 22 '13 at 22:36
  • By the way, it is also possible to upvote an answer by clicking on the arrow on the left :-) – Stefan Hamcke Sep 22 '13 at 22:38