Real Analysis Exam Questions: Need Help Studying!

[Kuzu]

Real Analysis Exam Questions. Please Help!

I'm taking this course on real analysis and my exam will be in less than a week from now
These are exam questions from previous year which have been assigned as homework, and I just started studying and it's really hard. I would be sooo happy if anyone could help me with these question. Some of them will probably be asked in the exam...
It would be great if you can help me solve any of these. Thanks very much!

 

[micromass]

Wow, I pretty think this forum isn't for that. You can't just expect us to solve all your questions just like that.

You can however ask us some questions you really don't understand. We will help you then, IF you shown us some effort from your side.

I think this post was a bit rude of you, I'm sorry...

 

[Kuzu]

Oh sorry, thanks for being honest about my post, you are so right about that. I realize how rude it seems but I'm really desperate about these questions and couldn't make much progress solving them. I'm working on them... will post here about everything I come up.

I thought maybe someone could help me. It's a big favor I know. Sorry again for doing that :(

 

[micromass]

I understand that you're desperate and I really want to help you with your questions. Butwe can't solve them all just like that

Which one are you struggling the most with?

 

[Kuzu]

Hey thank you,
I made some progess with a few questions but I still have big problems with some of them.

For example with question 2.
I guess to show that it is not countable, I have to suppose it is countable and arrive at a contradiction.
For a set to be countable there should be a way for listing them.
So I should be able to put it into one-to-one correspondence with the natural numbers.

I tried to sketch the list but there are infinitely many point between any two points I choose. Thus I understand its not countable but how to show it mathematically?

If there are infinately many points in this set it means to me infinite, so no way to count them, but from my notes I see the definition that any countable set is infinite.
This is exactly the opposite of what makes sense to me, I'm so confused :(

Can you help me?

 

[micromass]

Hmm, what did you prove in your course regarding countability?
Did you show that \mathbb{R} is uncountable? Or that [0,1] is uncountable?

Also, the word countable might not be the best term. Countable does not mean you can count how many objects are in the set. Otherwise only finite sets would be countable.

Informally, countability means that you can perform the following process: choose an element of the set, choose a second element of the set, choose a third element of the set, and so on. If you chose them right, then you will have exhausted the set.

Example: in \mathbb{N}, you can choose 0 as first element, 1 as second element, ..., n as nth element. After a while, I have chosen all elements in my set. (Note that I could also choose 0 as first element, 2 as second element, ..., 2n as nth element. Then I would not have ended up with N. That's what I meant with "choose your elements right").

Forgive me if you don't understand much of this. But I wanted to make clear that "countability" and "being able to count how many elements are in the set" are not the same thing.

Anyway, for the question. What is in your course regarding countability?

 

[Kuzu]

Yes we have seen that R is uncountable and finite.
And N is countable and infinite.

I'm trying to make sense of all this. Thanks to you I think I got the idea a little now.
Ok so I will not think of it as counting infinitely many elements but being able to pinpoint the next element in the set. With N I can say 5 comes after 4. but in R I can't say 0.5 is after 0.4 because there is also 0.40001 after 0.4 and so on... So [0,1] in R is not countable in that sense.

Same thing for [0,1]x[0,1], but how can I prove this mathematically?

Thanks for your big help!

 

[Kuzu]

I found something called Cantor's Diagonal Argument.
If I say [0,1] subset of R is countable and make a list and then construct a new number with cantor's argument to prove [0,1] is uncountable, can I then say [0,1]x[0,1] subset of R^2 is also uncountable? Is this a correct solution?

 

[micromass]

Yes, it's good to have some informal intuition about countable sets. But the way of viewing countability as not-being-able-to-pinpoint-next-element is just an intuition, I don't think the professor will like it that you use it in formal proofs

So, we will have to be rigurous about this one. We know that R is uncountable. The correspondance \tan:]-\pi/2,\pi/2]\rightarrow \mathbb{R} is bijective and thus ]-\pi/2,\pi/2 [ is also uncountable. With a similar function, we can show that ]0,1[ is uncountable.

This implies that [0,1] is uncountable (this is formally proven by Cantor-Bernstein, but I don't know if you've seen this). This is true since [0,1] contains the uncountable subset ]0,1[. So [0,1] also has to be uncountable. (If you have problems proving this explicitly, say it (: ).

Now for [0,1] x [0,1]. There exists an injection

f:[0,1]\rightarrow [0,1]\times [0,1]:x\rightarrow (x,x)

so in some sense [0,1] has the same cardinality as a subspace of [0,1]x[0,1]. Since this subspace is uncountable, the bigger space [0,1] x [0,1] must be uncountable.

Tell me if you have troubles with formalizing this arguments. But it would already be a lot if you understand 90% of what I've written...

 

[micromass]

Cantors diagonal argument could work to (: but beware, the proof seems easy, but there are some pitfalls

 

[JonF]

once you get [0,1] is uncountable [0,1]x[0,1] follows immediately since [0,1] can be trivially bijectivly mapped to [0,1]x{0}. Which would be an uncountable subset of [0,1]x[0,1]

 

[Kuzu]

Micro thanks so much!
We didn't cover Cantor-Berstein but I think I get the idea.. I will consult my teacher about this method and if I can use it.

btw is it ok to say that [0,1] has same cardinality as the whole set [0,1]x[0,1]?

Thanks to you too JonF,
But can't I just say that if [0,1] is an uncountable subset of [0,1]x[0,1] than it too has to be uncountable?

I think I'll be able to answer this question now :) thanks guys


About question 4.
I got the only limit point to be 0 so A'={0}
from there \overline{A}=AuA'= A itself,
and A^o={}
is this correct?

what about the second part of the question?
how can I construct a subset of R with a countable set of limit points?
isn't set A already like that because set of limit points {0} is a countable?

or could I construct B={0}u{1/n:neN}u{1+1/n:neN}<R which would have limit points 0 and 1?

sorry I'm not familiar with the latex code

 

[micromass]

Yeah, ask your professor if you could use Cantor-Bernstein. If not, say something, maybe I can devise a proof which does not use it...

Yes, it is ok to say that [0,1] has the same cardinality as [0,1]x[0,1]. But the proof of this is not easy. And I don't think you will need to be able to prove it...

Question 4 is correct.
To construct the set with countably many limit points. You're on the right way:
\{0\}\cup\{1/n~\vert~n\in \mathbb{N}\} has 0 as limit point
\{0\}\cup\{1/n~\vert~n\in \mathbb{N}\}\cup\{1+1/n~\vert~n\in \mathbb{N}\} has 0 and 1 as limit points.
\{0\}\cup\{1/n~\vert~n\in \mathbb{N}\}\cup\{1+1/n~\vert~n\in \mathbb{N}\}\cup\{2+1/n~\vert~n\in \mathbb{N}\} has 0,1 and 2 as limit points.

Can you now come up with a set which has \mathbb{N} as limit points? Just use a modification of above...
If you can answer this, then question 7 shouldn't be to hard either...

 

[Kuzu]

Ok thanks I'll do that, you are an angel :)

Hmm, for 1000 limit points
something like this maybe? I'm not sure.. is there another way?

999
U{m+1/n|neN}
m=0


do I have to add union {0} also?

I'm so new to all this , and my exam is on monday.. I hope I'll finish these until then.
Now I have to go to sleep.. I'll study hard tomorrow and let you know how far I came.

I can't thank you enough..

 

[micromass]

Yes, that set has 1000 limit points. No, you don't need to add {0}.

Good luck on your studies. And... don't hesitate to ask if something is unclear, we're here to help you

 

[Kuzu]

Thanks and hi,
Yesterday I studied further topics and learned some new theorems in real analysis..
still trying to digest all the new definitions like compactness, open subcovers, boundedness.. :shy:

I tried to solve questions 3 and 5, but I'm not really sure if its all correct and I have big questions. Hoping you can help me..

For Q3 I found:

for 1. Part "Show that E is closed,bounded and compact in Z"
E is [-100,100] in Z<R , it is finite and contains no limit points, which means its closed. right?

Also E<N₁₀₁(x) (it has a single neighborhood big enough that contains all of E), so it is bounded.

but how can I show if its compact in Z? In R it would be because of Heine-Borel theorem. here I god a little stuck

What should I do with given metric d(x,y)=|x-y| ? How can I use it?

2. part "Is E open in Z?"
E cannot be open in Z because there exist N₁(100) that is not interior.

3. part "Is N a compact subset of Z?"

N is infinite so it can't be bounded, so N is not compact.

Is this all enough to prove? Should I write more formally? how?

For Q5,
As I understand, set X is the entire xy-axis together with the closed unit disc centered at 0.

Here this N(0) confuses me a little. for R² what is N(0)? is it the origin (0,0)? what would N(1) be?
I was not sure about this but I said for (0,0) which is also center of the unit disc..

1. part "Write all neighborhoods N_e(0)as e varies in (0,infty)"
Y = U_{e=0 to infty} N_e(0)

2. part "is subset A of X open,closed or compact?"
-It is no open because there exists xeX that is not an interior point (any point on the boundery of the disc)

-A contains all its limit points, so it is closed.

-The subset A has a finite cover such as {N₂(0)}, so it must be bounded.

And from Heine–Borel theorem I can say that because A is closed and bounded, it will be compact in Rⁿ.

Am I on the right foot here?


..btw I asked about Cantor-Berstein and I will have to use Cantor's Diagonal Argument which in fact we have seen in class. (I didn't know that, but still thanks for your example)

 

[JonF]

Thanks to you too JonF,
But can't I just say that if [0,1] is an uncountable subset of [0,1]x[0,1] than it too has to be uncountable?

[0,1] is not a subset of [0,1]x[0,1]

 

[micromass]

Q2: So this will be just like proving that [0,1] is uncountable. Except that now you will have to deal with 2 coordinates. Tell me if you experience problems with this.

For question 3 and 5, I would first like to make a general remark:
You have seen the Heine-Borel theorem. This states that a set is compact in R (or more general in Rⁿ) iff it is closed and bounded. The metric space involved was always R or Rⁿ. Right here you are dealing with subspaces of R, these are other metric spaces, so you can't apply Heine-Borel to this questions (at least not directly).

Example: in [0,1[, the set [1/2,1[ is closed and bounded but it is not compact. But we do have the following theorem:

Let X be a set of R, let Y\subseteq X. If Y is compact in R, then it is compact in X. (so if you want to know if Y is compact in X and if X is closed, apply Heine-Borel to see if it is compact in R and then apply this theorem).



Now to the separate questions:

Q3: you are correct that saying that this set is closed and bounded in Z. To see if the set is compact in Z, you'll need to use the above theorem. That requires you to check if the set is closed and bounded in R. Fortunately, this is the case.

The set E is in fact open in Z. For every x in E, we observe that N_{1/2}(x)\subseteq E (observe that N_{1/2}(x)=\{x\} in Z). So around every x of E, we can find an open ball which is still in E. This proves that every point x is an interior point.
Note that, in fact, EVERY SET is open in Z. The same prove as above applies.

N is indeed not compact in Z. Your proof of this is also fine.

Q5: I think the first part of your question asks you to describe all balls around (0,0) with radius epsilon. Methinks a sketch of this will be fine.

A is indeed not open. But beware that your proof of this has a few holes. You say that any point on the boundary of the disk isn't an interior point. This is false. The points {(1,0),(-1,0),(0,1),(0,-1)} indeed arent interior points, but the rest are! So a good proof would be to say that (for example) (1,0) isn't an interior point.

the set is indeed closed and bounded in X. But again, you can't use Heine-Borel just yet, you'll have to show that it is closed and bounded in R. This is in fact the case (by virtually the same proof), thus the set is compact in R. And therefore also compact in X.



Btw, when is your exam?

 

[Kuzu]

[0,1] is not a subset of [0,1]x[0,1]

Oh sorry, I thought subset and subspace were the same. Thanks
So [0,1] is subspace of [0,1]x[0,1] right?

Q2: So this will be just like proving that [0,1] is uncountable. Except that now you will have to deal with 2 coordinates. Tell me if you experience problems with this.

This looks hard in 2D, can't I just prove [0,1] with Cantor's Diagonal and then show [0,1]x[0,1] can't be countable either?
By showing injection and cardinality..

Now for [0,1] x [0,1]. There exists an injection

f:[0,1]\rightarrow [0,1]\times [0,1]:x\rightarrow (x,x)

so in some sense [0,1] has the same cardinality as a subspace of [0,1]x[0,1]. Since this subspace is uncountable, the bigger space [0,1] x [0,1] must be uncountable.

About Heine-Borel, thanks for pointing that out. I will be more carefull from now on and use your suggested theorem together with H-B .

Q3: I thought I can only look for neightborhoods such as 1,2,3... not 1/2 because we are in Z. That was my fault. thanks!

btw you say every set is open in Z, but we said E is finite,closed and bounded in Z. sorry there is something I don't understand. In part 1 we show E is closed and in part 2 we show E is open, both in Z.


Q5: Oh yes only points {(1,0),(-1,0),(0,1),(0,-1)} are not interior. thanks!


About Q6.

Should I apply metric rules?

for d(x,y) = |x2-y2| in metric space (0,inf)

1)d(x,y) >= 0 is true
2)d(x,y) = 0 if x=y is true
3)d(x,y) = d(y,x) truee
4)d(x,z) <= d(x,y) + d(y,z) also true

so this is a metric for that metric space.

For part 2: Is it ok just to say U_{e=0 to inf}{N_e(3)} Is the answer this simple?



btw. I found some good videos watching them right now. ()

My exam is in 16 hours from now :) I don't feel ready but at least I made some effort and will be able to solve some part..
In class I don't understand much because its fast and I didn't know anything when I posted this thread. Now I definitely think I have some chance..
btw I'm from Turkey, sorry for my funny english :)

 

[JonF]

Oh sorry, I thought subset and subspace were the same. Thanks
So [0,1] is subspace of [0,1]x[0,1] right?

ack subset and subspace aren't the same! and this is neither. If it's a subspace it's a subset. The converse isn't necessarily true. A subspace is a subset that preserves what ever defines the space your talking about.

Ask yourself what do elements of [0,1]x[0,1] look like, write a few down.
What do elements of [0,1] look like? Write a few down.

Can any element of [0,1] be in [0,1]x[0,1]?

Now write a few down from [0,1]x{0}, or [0,1]x{1/2,1/3,1/4}

 

[micromass]

This looks hard in 2D, can't I just prove [0,1] with Cantor's Diagonal and then show [0,1]x[0,1] can't be countable either?
By showing injection and cardinality..

Well I suppose you could do that. But perhaps your professor wants you to prove the uncountability of [0,1]x[0,1] directly.
This is done by a simple modification of Cantor's Diagonal Argument:
Assume that [0,1]x[0,1] is uncountable, then we can find a bijection f:N-->[0,1]x[0,1]. Then
f(1)=(0,x_{1,1}x_{1,2}x_{1,3}...; 0,y_{1,1},y_{1,2},y_{1,3}...)
f(2)=(0,x_{2,1}x_{2,2}x_{2,3}...; 0,y_{2,1},y_{2,2},y_{2,3}...)
f(3)=(0,x_{3,1}x_{3,2}x_{3,3}...; 0,y_{3,1},y_{3,2},y_{3,3}...)
But now we set u_1=x_{1,1}+1, u_2=x_{2,2}+1, u_3=x_{3,3}+1 (where I put 9+1=0). Similarly, we sey v_1=y_{1,1}+1, v_2=y_{2,2}+1, v_3=y_{3,3}+1. Then
(0,u_1u_2u_3... ; 0,v_1v_2v_3...)
can not be written as an f(n). Thus f is not surjective.

btw you say every set is open in Z, but we said E is finite,closed and bounded in Z. sorry there is something I don't understand. In part 1 we show E is closed and in part 2 we show E is open, both in Z.

Remember this for the future: a set can be open and closed in the same time! In R, the only sets which are open and closed are the empty set and R. But in Z, every set is open and closed.

About Q6.

Should I apply metric rules?

for d(x,y) = |x2-y2| in metric space (0,inf)

1)d(x,y) >= 0 is true
2)d(x,y) = 0 if x=y is true
3)d(x,y) = d(y,x) truee
4)d(x,z) <= d(x,y) + d(y,z) also true

so this is a metric for that metric space.

Yes.

For part 2: Is it ok just to say U_{e=0 to inf}{N_e(3)} Is the answer this simple?

I'm afraid not. Your instructor probably wants to see explicitly what the open sets are. I mean, you should probably be able to draw these sets.

Good luck on your exam!

 

[Kuzu]

\infty<-- Don't bother this, latex issue...


Hey! I just returned from the exam.. and it was really good!

There were two questions almost the same as
Q3 with little change x²<=1000000, so the set is [-1000,1000] and everything else the same.

Q4 with addition to show Boundery \partialA=A-A^o=?
I didn't understand it exactly but I said \partialA = A-A^o = A - {} = A
other than that everything was full and correct.

I also did all other questions, probably all right. They didn't ask [0,1]x[0,1] but I'm totally ready for such question for 2nd midterm exam :)


Thanks again for your help!

 

[micromass]

Wow, congratulations! I'm really happy for you