Show Cardinality of Real Numbers and Complements

[Mr Davis 97]

Homework Statement

$\mathbb{R} \setminus C \sim \mathbb{R} \sim \mathbb{R} \cup C$.

Homework Equations

The Attempt at a Solution

I have to show that all of these have the same cardinality. For $\mathbb{R} \cup C \sim \mathbb{R}$, if $C = \{c_1, c_2, ... c_n \}$ is finite we can define $f(x) = \begin{cases} n &~ \text{if} ~x=c_n~ \text{where} ~n \le |C| \\ n+|C| &~ \text{if} ~x=n~ \text{where} ~n \in \mathbb{N} \\ x &~ \text{if} ~x \not\in \mathbb{N} \cup C \end{cases}$

And if $C = \{c_1, c_2, c_3, ... \}$ is infinite we can define
$f(x) = \begin{cases} 2n &~ \text{if} ~x=c_n~ \text{where} ~n \in \mathbb{N} \\ 2n-1 &~ \text{if} ~x=n~ \text{where} ~n \in \mathbb{N} \\ x &~ \text{if} ~x \not\in \mathbb{N} \cup C \end{cases}$

I think that these are both bijections.

However, I am a little confused about showing that $\mathbb{R} \sim \mathbb{R} \setminus C$. Does this imply that all the elements of $C$ are in $\mathbb{R}$? If this is the case couldn't we just define $f: \mathbb{R} \rightarrow \mathbb{R} \setminus C$, and have a very similar bijection as the first one defined above?

 

[mfb]

How is C defined? It has to be defined somewhere, don't leave that out.

What exactly are your functions f supposed to be? They look like they map real numbers to real numbers, but if all elements of C are real numbers, then there is a trivial bijection.

 

[Mr Davis 97]

How is C defined? It has to be defined somewhere, don't leave that out.

What exactly are your functions f supposed to be? They look like they map real numbers to real numbers, but if all elements of C are real numbers, then there is a trivial bijection.

Sorry, I forgot an important piece of information. The only information given on $C$ is that it is countable, so either finite or denumerable.

 

[mfb]

Then I guess C can contain both real numbers (interesting for R\C) and other elements (interesting for R u C).

 

[Mr Davis 97]

Then I guess C can contain both real numbers (interesting for R\C) and other elements (interesting for R u C).

So is what I have above correct at least for showing that $\mathbb{R}$ and $\mathbb{R} \cup C$ have the same cardinality?

 

[mfb]

See my question above: What is f? You didn't introduce it properly.

 

[Mr Davis 97]

See my question above: What is f? You didn't introduce it properly.

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

 

[mfb]

What happens if c₂=3? Then you assign two different values to f(3).
The approach can work, but it needs a bit more refinement.

 

[Mr Davis 97]

Let $C' = C \setminus \mathbb{N}$

If $C' = \{c_1, c_2, ... c_n \}$ is finite we can define $, f(x) = \begin{cases} n &~ \text{if} ~x=c_n~ \text{where} ~n \le |C'| \\ n+|C| &~ \text{if} ~x=n~ \text{where} ~n \in \mathbb{N} \\ x &~ \text{if} ~x \not\in \mathbb{N} \cup C' \end{cases}$

And if $C' = \{c_1, c_2, c_3, ... \}$ is infinite we can define
$f(x) = \begin{cases} 2n &~ \text{if} ~x=c_n~ \text{where} ~n \in \mathbb{N} \\ 2n-1 &~ \text{if} ~x=n~ \text{where} ~n \in \mathbb{N} \\ x &~ \text{if} ~x \not\in \mathbb{N} \cup C' \end{cases}$

Does that work?

 

[mfb]

That should be fine.

 

[Mr Davis 97]

That should be fine.

How would I do the set difference one? I would use a bijection similar to the ones I used above, but I am concerned with the fact that we are using a set difference. Usually if we have $f: A \cup B \rightarrow A$ and we are trying to show that $A \cup B$ has the same cardinality as $A$, we would take the extra elements of $A \cup B$ and embed them in a denumerable subset of $A$. However, since we are using set difference in this case, if I choose that denumerable subet to be the natural numbers, it seems that there is no reason that C could be that same denumerable set (meaning that the natural numbers is no longer a denumerable set of $\mathbb{R} \cup C$. So I am not sure how to address this.

 

[mfb]

How would I do the set difference one?

You can do it very similarly to the other one. Basically reversing the direction.