# Define the bijection

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

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#### dwsmith

MHF Hall of Honor
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 $$\displaystyle x \neq \frac {1}{2^n}$$ 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....?

we r in reals...

#### dwsmith

MHF Hall of Honor
Are you trying to show (0,1) has cardinality c?

#### undefined

MHF Hall of Honor
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 $$\displaystyle x \neq \frac {1}{2^n}$$ 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.)

novice and Swlabr

#### novice

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

MHF Hall of Honor
$$\displaystyle f: (0,1) \rightarrow [0,1]$$<---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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novice

#### novice

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.

#### novice

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

MHF Hall of Honor
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 $$\displaystyle x \neq \frac {1}{2^n}$$ 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 $$\displaystyle \gamma+\kappa=\gamma$$