Cardinal arithmetic

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  • #26
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Well.
first, what is the difference?
there is some,
there exists,
i cant quite make the difference, could you give me an example of where one is true and the other is not?

moving on,
there is some ##y \in X##such that## f(y) = m## and if ## m = n +1 ## then g(x) = m by the definition of ##g##.
but from the definition of ##g, y \in X \cup \{ x \} ## so y is an element of X or {x}. surely that is enough evidence to conclude that there exists ## y \in X \cup \{ x \} ## such that g(y) = m... or is this a case of there is some?
 
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  • #27
LCKurtz
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No, you are still not getting it. I am not arguing about the semantic difference between saying "there is some ##y##" and "there exists ##y##". I am trying to get you to understand the proof of that statement. You stated in your argument in post #24 that ##g(y) = f(y) = m## for all ##y\in X##. That is a false statement. In post #25 I explained what your statement should have been and asked whether you see the difference. Do you understand the difference?
 
  • #28
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I see the difference. to state that for all ## y \in X, f(y) = m ## is a false statement as it implies that the image of every y in X is m. I understand the difference. I do apologise i wrongfully assumed that you were trying to point out some apparent difference between 'there exists' and 'there is some'. Thanks for pointing that one out!

To complete the proof would the second half of my last post be sufficient? i seem to think so, because in the case y is in X then there exists a bijection and in the case y is in {x} we have a map g(x) = m.
 
  • #29
LCKurtz
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ok, now to prove the 'there exists' part;
let ## m \in S_{n+1}## we must show that there exists ##y \in X \cup \{x\}## such that ##g(y) = m##


moving on,
there is some ##\color{red}{y \in X}## such that ## \color{red}{f(y) = m}## and if ## m = n +1 ## then g(x) = m by the definition of ##g##.
but from the definition of ##g, y \in X \cup \{ x \} ## so y is an element of X or {x}. surely that is enough evidence to conclude that there exists ## y \in X \cup \{ x \} ## such that g(y) = m... or is this a case of there is some?

To complete the proof would the second half of my last post be sufficient? i seem to think so, because in the case y is in X then there exists a bijection and in the case y is in {x} we have a map g(x) = m.

You are close to showing ##g## is onto, but not quite done. The top quote shows what you are trying to prove to show ##g## is onto. What I have highlighted in red is what you have shown. You have ##f(y)=m## and you need ##g(y)=m##. Like I said before, it is easy but you need to explain how you get that step.

Please do that next.

With respect to showing ##g## is one-to-one, instead of trying to discuss all the preceding posts, I would prefer you give a complete argument in a following post. Carefully state what you are trying to prove and give your argument to prove it, and I will discuss your argument with you.
 
  • #30
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I am setting out to prove that g is onto;
let ## m \in S_{n+1} ##, we must show that there exist ## y \in X \cup \{x\}## such that ## g(y) = m ##.

since ##f## is a bijection between X and ##S_n## there is some ##y \in X ## such that ##f(y) = m## and if ##m= n+1 ## then ##g(x) = m ## by the definition of g ALSO by the definition of ##g##, ##y \in X \cup \{x\}## this implies that ##y \in X## or ## y \in \{x\}##, since ##x \in \{x\}## we have ## x,y \in \{x\}## this implies that if ##m = n+1##, then ##g(x) = g(y) = m##.


Here i have show that there exist ## y \in X \cup \{x\}## such that ## g(y) = m ##

You have to understand, i have run out of ideas. RUN out of ideas.

I had an idea earlier but i cant seem to do anything with it, using the fact that m = n + 1, that is m - 1 = n...
 
  • #31
LCKurtz
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I am setting out to prove that g is onto;
let ## m \in S_{n+1} ##, we must show that there exist ## y \in X \cup \{x\}## such that ## g(y) = m ##.

since ##f## is a bijection between X and ##S_n## there is some ##y \in X ## such that ##f(y) = m##

That is only true if ##m\le n##, not if ##m## happens to be ##n+1##. So state that in your argument.

and if ##m= n+1 ## then ##g(x) = m ## by the definition of g

That is the other case

ALSO by the definition of ##g##, ##y \in X \cup \{x\}## this implies that ##y \in X## or ## y \in \{x\}##, since ##x \in \{x\}## we have ## x,y \in \{x\}## this implies that if ##m = n+1##, then ##g(x) = g(y) = m##.

I guess that is a very confusing way to say the domain of ##g## is ##X\cup \{x\}## and ##g(x) = m+1##. You already said that above "by the definition of g".
Here i have show that there exist ## y \in X \cup \{x\}## such that ## g(y) = m ##

No you haven't. You have shown that if ##m=n+1## that ##g(x)=m##. And you have shown that if ##m\le n## there is a ##y\in X## such that ##\color{red}{f}(y) = m##. YOU HAVE TO SHOW THAT ##\color{red}{g}(y) = m##.
You have to understand, i have run out of ideas. RUN out of ideas.

Perhaps it is time to print out this thread, take it to your teacher, and go over it with him/her in person. That would be a much more efficient use of time.
 
  • #32
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Actually i don't have a teacher/ lecturer, i made the choice to drop out of university to look after a dying grandparent, this resulted in me taking a degree with the open university which does not cover analysis of this kind. so i am self-studying Analysis 1 by terrence tao. When i do come across a problem i have to post it on here. I have NO other choice. I don't have a lecturer to ask.

It's not that im not applying myself, i spend countless hours on this book.

But thanks for your time
 
  • #33
LCKurtz
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OK. Then respond to my last post. You have to show ##g(y)=m##. What is ##g(y)## according to its definition?
 
  • #34
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for any ##y \in X \cup \{x\}##
we define;
##g(y):= f(y)## for all ## y \in X##,
##g(x):=n+1 ##
 
  • #35
LCKurtz
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for any ##y \in X \cup \{x\}##
we define;
##g(y):= f(y)## for all ## y \in X##,
##g(x):=n+1 ##

So tell me how you know there is a ##y## such that ##g(y)=m##. I know, it's easy. But you have to say why.
 
  • #36
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Because y is an element of the set X U {x}
 
  • #37
LCKurtz
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No. It is because if ##m\le n##, so ##m\in S_n##, you have there is ##y\in X## such that ##f(y)=m## because ##f## is a bijection between ##X## and ##S_n##. And by the definition of g, if ##y\in X## then ##g(y)=f(y)##. That is why ##g(y) = m##.

This should complete the argument that ##g## is onto ##S_{n+1}##. Now, since we have been through it and it isn't really a homework problem because you aren't actually taking a class, I'm going to stretch the PF rules and show you how you should write it up. We have called$$
S_n = \{ i\in N: i\le n\}$$You are given a bijection ##f## from a set ##X## to ##S_n##. If ##x## is not an element of ##x##, you want to demonstrate a bijection ##g## from ##X\cup \{x\}## to ##S_{n+1}##.

Our proposed bijection is$$
g(y)=\left\{\begin{array}{rl}
f(y),& y\in X\\
n+1,& y=x
\end{array}\right.$$
Claim: ##g## is onto ##S_{n+1}##
Proof: Suppose ##m\in S_{n+1}##. If ##m = n+1## then ##g(x) = n+1=m##. If ##m < n+1## then ##m \in S_n## so, since ##f## is a bijection between ##X## and ##S_n##,(hence onto), there is a ##y\in X## such that ##f(y) = m##. Since ##y\in X## then ##g(y) = f(y)## by the definition of ##g##, so ##g(y)=m##.
So in each case, ##m=n+1## or ##m\le n## we have found a ##y## such that ##g(y) = m##.

Now we must show that ##g## is 1-1. You do it. Write down carefully what you have to prove and argue it.
 
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  • #38
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Thank you ever so much.

The first line
If m = n+1 then g(x) = m+1. I find this a little confusing. if m is some element of ##S_{n+1}## and m = n+1, i get this part.... but then to say that g(x) = m+1.
This may sound really stupid, but to me, i'm not quite seeing it.

Claim: ##g## is a 1-1 function
Proof: Let ##m \in S_{n+1} ## and ##y_1, y_2 \in X \cup \{x\} ##. suppose that ##g(y_1) = g(y_2) ##. Then we have two cases to deal with, if ## m < n ## then by the definition of ##g##, ##g(y_1) = g(y_2) ## becomes ##f(y_1) = f(y_2) ##, but ##f## is a bijection, so it follows that ##y_1=y_2##, in the second case, if ## m = n +1 ## then ##g(y_1) = g(y_2)## can only occur if ##y_1 = y_2 = x ##, and##g(x) = m+1##. In either case, g maps different elements to different elements. It follows that ##g## is 1-1
 
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  • #39
LCKurtz
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Thank you ever so much.

The first line
If m = n+1 then g(x) = m+1. I find this a little confusing. if m is some element of ##S_{n+1}## and m = n+1, i get this part.... but then to say that g(x) = m+1.
This may sound really stupid, but to me, i'm not quite seeing it.

Not stupid and you shouldn't see it. That was just a typo. I fixed it to ##g(x) = n+1##.

Claim: ##g## is a 1-1 function
Proof: Let ##m \in S_{n+1} ## and ##y_1, y_2 \in X \cup \{x\} ##. suppose that ##g(y_1) = g(y_2) \color{red}{=m}##. Then we have two cases to deal with, if ## \color{red}{m < n} ## then by the definition of ##g##, ##g(y_1) = g(y_2) ## becomes ##f(y_1) = f(y_2) ##, but ##f## is a bijection, so it follows that ##y_1=y_2##, in the second case, if ## m = n +1 ## then ##g(y_1) = g(y_2)## can only occur if ##y_1 = y_2 = x ##, and ##g(x) = m+1##. In either case, g maps different elements to different elements. It follows that ##g## is 1-1

Pretty good. You have to say what ##m## is, which I fixed, and is the inequality in red should read ##m\le n##.

Have a nice holiday.
 
  • #40
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Many thanks again, you also! :)
 

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