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My Question reads:

Prove: if $A$ is denumerable, then $A\setminus\{x\}$ is denumerable.

I have proposed this:

Let $f:\mathbb{N}\to A$ be a bijection. Let $g:\mathbb{N}\to A\setminus\{x\}$ be a function defined by:

$$ g(y)= \begin{cases} y, &y=1\\ y+1, &y\geq 2 \end{cases} $$ I just wanted to see if the answer I have provided is a reasonable choice for a bijection. This is an answer I have come up with myself. I want a verification of my answer and/or clarification.

Sam
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    How is your function going to avoid $;x;$ in the image? – DonAntonio Dec 08 '16 at 21:45
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    This is a bijection from $\mathbb{N}$ to $A-{f(2)}$. – kccu Dec 08 '16 at 21:45
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    @Sam, you must take care your function is defined onto $;A\setminus{x};$ ... – DonAntonio Dec 08 '16 at 21:58
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    What I am saying is that your function $g$ is defined so that $g(1)=f(1)$, $g(2)=f(3)$, $g(3)=f(4)$, $...$, so the only element in $A$ that it "misses" is $f(2)$. What you want, instead, is a function that "misses" $x$. So your function is fine if $f(2)=x$, but in general that won't be the case. To fix it, note there exists $k \in \mathbb{N}$ such that $f(k)=x$. Now how do you define $g$ so it "misses" $f(k)$? – kccu Dec 08 '16 at 21:59

2 Answers2

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Using your idea:

Let $f: \mathbb N \rightarrow A$ be an appropriate bijection. Let $f(k) = x$

Let $g(n) = f(n)$ if $n < k$. Let $g(n) = f(n+1)$ if $n \ge k$.

It may be easier to see this as index rather thann functions (although they are exactly the same):

Let $A = \{a_i\}$ be an indexing for A then $x = a_k$ for some $k$. Let $b_i = a_i$ for all $i < k$ and let $b_i = a_{i+1}$ for all $i \ge k$. Then $\{b_i\} = A\setminus \{x\}$ and $\{b_i\}$ is an indexing that maps $\mathbb N \rightarrow A\setminus \{x\}$

===== thourough answer ====

$A$ is denumerable. That means we can write the terms of $A$ into a numbered list like so:

$1 \rightarrow f(1) = a_1 = w$

$2 \rightarrow f(2) = a_2 = \sigma$

$3 \rightarrow f(3) = a_3 = z$

.....

$k-1 \rightarrow f(k-1) = a_{k-1} = \psi$

$k \rightarrow f(k) = a_k = x$

$k + 1 \rightarrow f(k+1) = a_{k+1} = r$

$k + 2 \rightarrow f(k+2) = a_{k+2} = v$

.....

(I just made up those values $w, \sigma, z,$ etc. The idea being that those are the elements of $A$ and are .... whatever they are. You list them in order because... we can.)

Now we can simply remove the $k-th$ element $x$ and list the values as:

$1 \rightarrow f(1) = a_1 = w$

$2 \rightarrow f(2) = a_2 = \sigma$

$3 \rightarrow f(3) = a_3 = z$

.....

$k-1 \rightarrow f(k-1) = a_{k-1} = \psi$

empty line-- it's gone

$k + 1 \rightarrow f(k+1) = a_{k+1} = r$

$k + 2 \rightarrow f(k+2) = a_{k+2} = v$

.....

But now we are skipping over the matching of any $k-th$ term.

But that's okay. We'll just lower all the numbers by $1$ to make this list:

$1 \rightarrow f(1) = a_1 = w$

$2 \rightarrow f(2) = a_2 = \sigma$

$3 \rightarrow f(3) = a_3 = z$

.....

$k-1 \rightarrow f(k-1) = a_{k-1} = \psi$

empty line we skipped

$k \rightarrow f(k+1) = a_{k+1} = r$

$k + 1 \rightarrow f(k+2) = a_{k+2} = v$

.....

Basically we are creating a new indexing. $f(n)$ and $a_n$ is the old indexing and $g(n)$ and $b_n$ will be the new indexing.

$1 \rightarrow f(1) = a_1 = g(1) = b_1 = w$

$2 \rightarrow f(2) = a_2 = g(2) = b_2 =\sigma$

$3 \rightarrow f(3) = a_3 = g(2) = b_2 =z$

.....

$k-1 \rightarrow f(k-1) = a_{k-1} =g(k-1) = b_{k-1} = \psi$

empty line we skipped

$k \rightarrow f(k+1) = a_{k+1} = g(k) = b_k r$

$k + 1 \rightarrow f(k+2) = a_{k+2} = g(k+1) = b_{k+1} = v$

.....

To formalize we say: $A$ is denumerable so there is a bijection $f:\mathbb N \rightarrow A$. We can define $g:\mathbb N \rightarrow A$ via $g(n) = f(n)$ if $n < f^{-1}(x)$; $g(n) = f(n+1)$ if $n \ge f^{-1}(x)$. (For convenience, let's label $f^{-1}(x) = k$ i.e. $f(k) = x$.

We know this is a bijection as:

1) it is injective. If $g(j) = g(i)$ then: $g(j) = f(j')$ where $j' = j$ or $j' = j+1$ and $g(i) = f(i')$ where $i' = i$ or $i' = i +1$. But $f$ is injective so $j' = i'$. If $j < k$ then $j' = j = i' < k$. But if $i' < k$ then $i \le i' < k$and we have $i' = i$ so $j = i$.

If on the other hand $j \ge k$ and $j' = j+ 1 = i' \ge k + 1$. $i' \ge k+1$ so $i \ge i' -1 \ge k+1$ so $i' = i+1$ so $j' = i'1 = j + 1 = i + 1$. So $j = i$.

So $g$ is injective.

2) $g$ is surjective. If $w \in A \setminus \{x\}$ then $w \in A$ and there is some $j$ so that $f(j) = w$ and $j\ne k$ aos $w \ne x$.

If $j < k$ then $g(j)=f(j) = w$. If $j>k$ than $g(j) = f(j+1) = w$.

So for all $w$ there is a $j'$ where $g(j') = w$. So $g$ is surjective.

fleablood
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  • You seem to be getting your domain (N) and range (A) mixed up. x in A. x is not a natural number. But it is the the y-th item in the set A for some y. i.e. f(y) = x. So you want g(w) = f(w) if w < y and g(w) = f(w +1) if w >= y. It may be easier to see this as index. Let A = {a_i} and x = a_k.... – fleablood Dec 09 '16 at 00:53
  • So can define f as n in N such that f(n)=x? – Sam Dec 09 '16 at 01:01
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    Well, you don't define f as such. Because A is denumerable you know some bijection exist. Call it f. You know x is in A so there must be some number, called it n, so that f(n) = x. This nothing more or less than saying we can list all the elements of A in order and x is the n-th. Your prove is simply to say "we'll use the same mapping f(i)-> the i-th term for 1 through n-1, then we'll skip x which is the nth term and make the n+1th term the new nth term, and for all k >= n the new kth term, g(k), is going to be the old k+1th term which was f(k+1)". – fleablood Dec 09 '16 at 05:31
  • Sure... that's not any different than my first post (except for chose of what letters to use for variables) so I didn't really understand your question. – fleablood Dec 09 '16 at 19:00
  • Okay, no this wasn't a question this was just me stating what I have down. Thank you for the help, I get it now seeing it all out. – Sam Dec 09 '16 at 19:18
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As an extension to Jack's answer, you could prove that given $n\in\mathbb{N}$ there is a bijection between $\mathbb{N}$ and $\mathbb{N} \setminus \{n\}$. Let's call $g = f^{-1}$, then you can biject $\mathbb{N}$ and $\mathbb{N} \setminus \{g(x)\}$.

If you then show there is a bijection between $A \setminus \{x\}$ and $\mathbb{N} \setminus \{g(x)\}$ (knowing that $g: A\rightarrow\mathbb{N}$ is bijective), then you're done.

chelivery
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