Define the bijection

Sep 2009
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Define a bijection between (0,1) and [0,1].



I know that in order to prove this is to use a piecewise function. So this is what I have.
(This is a piecewise function just could not figure out how to put it on here)


This is what I did... Please help me if im wrong.

f(x) = x if \(\displaystyle x \neq \frac {1}{2^n} \) for any \(\displaystyle n \epsilon N \)​
f(x) =\(\displaystyle \frac{1}{2^n} \) if \(\displaystyle n \epsilon N\cup {0} \)
 
Last edited:

dwsmith

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Define a bijection between (0,1) and [0,1].



I know that in order to prove this is to use a piecewise function. So this is what I have.
(This is a piecewise function just could not figure out how to put it on here)


This is what I did... Please help me if im wrong.

f(x) = x if \(\displaystyle x \neq \frac {1}{2^n} \) for any \(\displaystyle n \epsilon N \)​
f(x) =\(\displaystyle \frac{1}{2^n} \) if \(\displaystyle n \epsilon N\cup {0} \)
Are we in Reals, Integers, Rationals....?
 

dwsmith

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Are you trying to show (0,1) has cardinality c?
 

undefined

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Define a bijection between (0,1) and [0,1].



I know that in order to prove this is to use a piecewise function. So this is what I have.
(This is a piecewise function just could not figure out how to put it on here)


This is what I did... Please help me if im wrong.

f(x) = x if \(\displaystyle x \neq \frac {1}{2^n} \) for any \(\displaystyle n \epsilon N \)​
f(x) =\(\displaystyle \frac{1}{2^n} \) if \(\displaystyle n \epsilon N\cup {0} \)
It helps if you specify the domain and codomain, as in

\(\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}\)

I think what you were going for is something like

\(\displaystyle f: (0,1) \rightarrow [0,1]\)

defined piecewise (I also haven't mastered the curly brace thing to make it look nice):

\(\displaystyle f(x) = 0\) if \(\displaystyle x = \frac{1}{2}\)

\(\displaystyle f(x) = 1\) if \(\displaystyle x = \frac{1}{4}\)

\(\displaystyle f(x) = \frac{1}{2^{n-2}}\) if there exists \(\displaystyle n > 2, n \in \mathbb{Z}\) such that \(\displaystyle x = \frac{1}{2^n}\)

\(\displaystyle f(x) = x\) if there does not exist \(\displaystyle n \in \mathbb{Z}\) such that \(\displaystyle x = \frac{1}{2^n}\)

Now, can you prove that this is a bijection?

(A few edits.)
 
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Sep 2009
502
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It helps if you specify the domain and codomain, as in

\(\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}\)

I think what you were going for is something like

\(\displaystyle f: (0,1) \rightarrow [0,1][/math]<---we have a problem here; namely, \(\displaystyle 0 \notin (0,1)\)

defined piecewise (I also haven't mastered the curly brace thing to make it look nice):

\(\displaystyle f(x) = 0\) if \(\displaystyle x = \frac{1}{2}\)

\(\displaystyle f(x) = 1\) if \(\displaystyle x = \frac{1}{4}\)

\(\displaystyle f(x) = \frac{1}{2^{n-2}}\) if there exists \(\displaystyle n > 2, n \in \mathbb{Z}\) such that \(\displaystyle x = \frac{1}{2^n}\)

\(\displaystyle f(x) = x\) if there does not exist \(\displaystyle n \in \mathbb{Z}\) such that \(\displaystyle x = \frac{1}{2^n}\)

Now, can you prove that this is a bijection?

(A few edits.)\)
\(\displaystyle

Some times a function mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) can be bijective, but it seems to me that it's quite impossible to know what smallest real number \(\displaystyle x\) could go into \(\displaystyle f(x)\), let alone finding a bijection.

By definition, every element in the domain of a relation R must have an image in its codomain in order to be a function. In this case, we have no way of insuring every element in the domain to have an image since we don't even have the smallest element in it.\)
 

undefined

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\(\displaystyle f: (0,1) \rightarrow [0,1]\)<---we have a problem here; namely, \(\displaystyle 0 \notin (0, 1)\)
Sorry I don't follow. \(\displaystyle 0 \notin (0, 1)\) doesn't pose a problem? I never try to define \(\displaystyle f(0)\).

Some times a function mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\) can be bijective, but it seems to me that it's quite impossible to know what smallest real number \(\displaystyle x\) could go into \(\displaystyle f(x)\), let alone finding a bijection.

By definition, every element in the domain of a relation R must have an image in its codomain in order to be a function. In this case, we have no way of insuring every element in the domain to have an image since we don't even have the smallest element in it.
I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\), we are mapping from a subset of \(\displaystyle \mathbb{R}\) to a subset of \(\displaystyle \mathbb{R}\).

Would you mind re-reading my post and seeing if you still think it's problematic?

Edit: By the way, I wrote \(\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}\) as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like \(\displaystyle f: \{1,2,5,7,9\} \rightarrow \mathbb{C}\).
 
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Sep 2009
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Sorry I don't follow. \(\displaystyle 0 \notin (0, 1)\) doesn't pose a problem? I never try to define \(\displaystyle f(0)\).



I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\), we are mapping from a subset of \(\displaystyle \mathbb{R}\) to a subset of \(\displaystyle \mathbb{R}\).

Would you mind re-reading my post and seeing if you still think it's problematic?

Edit: By the way, I wrote \(\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}\) as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like \(\displaystyle f: \{1,2,5,7,9\} \rightarrow \mathbb{C}\).
I am really not an expert in functions. You seem to know more than I do. I am very glad to know that we need not be concerned over the smallest element in the domain. I need some time to think about your proposal.
 
Sep 2009
502
39
Sorry I don't follow. \(\displaystyle 0 \notin (0, 1)\) doesn't pose a problem? I never try to define \(\displaystyle f(0)\).



I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \(\displaystyle \mathbb{R}\) to \(\displaystyle \mathbb{R}\), we are mapping from a subset of \(\displaystyle \mathbb{R}\) to a subset of \(\displaystyle \mathbb{R}\).

Would you mind re-reading my post and seeing if you still think it's problematic?

Edit: By the way, I wrote \(\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}\) as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like \(\displaystyle f: \{1,2,5,7,9\} \rightarrow \mathbb{C}\).
Yes, you are right, we don't need to know the smallest element. I see \(\displaystyle n>2\).
 

Drexel28

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Define a bijection between (0,1) and [0,1].



I know that in order to prove this is to use a piecewise function. So this is what I have.
(This is a piecewise function just could not figure out how to put it on here)


This is what I did... Please help me if im wrong.

f(x) = x if \(\displaystyle x \neq \frac {1}{2^n} \) for any \(\displaystyle n \epsilon N \)​
f(x) =\(\displaystyle \frac{1}{2^n} \) if \(\displaystyle n \epsilon N\cup {0} \)
I know this does not answer your question, but it is true in general that if \(\displaystyle E\subseteq U\) is infinite then \(\displaystyle \text{card }E=\text{card }E\cup\{e_1,\cdots,e_n\}\)

In fact, if \(\displaystyle \gamma\) is an infinite cardinal number and \(\displaystyle \kappa<\gamma\) then \(\displaystyle \gamma+\kappa=\gamma\)
 
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