1. ## Define the bijection

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)

f(x) = x if $x \neq \frac {1}{2^n}$ for any $n \epsilon N$
f(x) = $\frac{1}{2^n}$ if $n \epsilon N\cup {0}$

2. Originally Posted by tigergirl
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)

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

3. we r in reals...

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

5. Originally Posted by tigergirl
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)

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

$f: \mathbb{R} \rightarrow \mathbb{R}$

I think what you were going for is something like

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

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

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

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

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

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

Now, can you prove that this is a bijection?

(A few edits.)

6. Originally Posted by undefined
It helps if you specify the domain and codomain, as in

$f: \mathbb{R} \rightarrow \mathbb{R}$

I think what you were going for is something like

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

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

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

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

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

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

Now, can you prove that this is a bijection?

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

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

Originally Posted by novice
Some times a function mapping from $\mathbb{R}$ to $\mathbb{R}$ can be bijective, but it seems to me that it's quite impossible to know what smallest real number $x$ could go into $f(x)$, 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 $\mathbb{R}$ to $\mathbb{R}$, we are mapping from a subset of $\mathbb{R}$ to a subset of $\mathbb{R}$.

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

Edit: By the way, I wrote $f: \mathbb{R} \rightarrow \mathbb{R}$ 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 $f: \{1,2,5,7,9\} \rightarrow \mathbb{C}$.

8. Originally Posted by undefined
Sorry I don't follow. $0 \notin (0, 1)$ doesn't pose a problem? I never try to define $f(0)$.

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 $\mathbb{R}$ to $\mathbb{R}$, we are mapping from a subset of $\mathbb{R}$ to a subset of $\mathbb{R}$.

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

Edit: By the way, I wrote $f: \mathbb{R} \rightarrow \mathbb{R}$ 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 $f: \{1,2,5,7,9\} \rightarrow \mathbb{C}$.
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.

9. Originally Posted by undefined
Sorry I don't follow. $0 \notin (0, 1)$ doesn't pose a problem? I never try to define $f(0)$.

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 $\mathbb{R}$ to $\mathbb{R}$, we are mapping from a subset of $\mathbb{R}$ to a subset of $\mathbb{R}$.

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

Edit: By the way, I wrote $f: \mathbb{R} \rightarrow \mathbb{R}$ 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 $f: \{1,2,5,7,9\} \rightarrow \mathbb{C}$.
Yes, you are right, we don't need to know the smallest element. I see $n>2$.

10. Originally Posted by tigergirl
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)

f(x) = x if $x \neq \frac {1}{2^n}$ for any $n \epsilon N$
f(x) = $\frac{1}{2^n}$ if $n \epsilon N\cup {0}$
I know this does not answer your question, but it is true in general that if $E\subseteq U$ is infinite then $\text{card }E=\text{card }E\cup\{e_1,\cdots,e_n\}$
In fact, if $\gamma$ is an infinite cardinal number and $\kappa<\gamma$ then $\gamma+\kappa=\gamma$