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Background: This comes from the book: INTRODUCTION TO MATHEMATICAL PROOFS Charles E. Roberts, Jr. Indiana State University Terre Haute, USA A Transition to Advanced Mathematics Second Edition

A set $A$ is denumerable if and only if $A\sim \mathbb{N}$.

$A\sim B$ if and only if there is a one-to-one correspondence (bijection) from $A$ to $B$.

Theorem 7.15 - If $A$ is a denumerable set and $B$ is a finite set, the $A\cup B$ is a denumerable set.

Question:

Prove that if $A$ is a denumerable set and $B$ is a finite subset of $A$, then $A\setminus B$ is denumerable.

Attempted proof - Note that

$$\begin{align*} A\setminus B &= A\cap B^c\\ &= (A\cup B)\cap B^c \end{align*}$$

Borrowing from the comments below. Since $A\cup B$ is denumerable by theorem 7.15, and any subset of a denumerable set is denumerable this implies that $A\setminus B$ is denumerable.

We know from theorem 7.15 that $A \cup B$ is denumerable. I am just not sure how to show that the complement of the finite set $B$ with $A\cup B$ is also denumerable.

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    $\begingroup$ Can't you just use the fact that subsets of denumerable sets are denumerable? $\endgroup$ – Don Thousand Mar 12 '19 at 4:03
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    $\begingroup$ Note that $A = (A\setminus B)\cup B$. What happens if $A\setminus B$ is not denumerable? $\endgroup$ – Corrêa Mar 12 '19 at 4:06
  • $\begingroup$ Other approach is to use the Hilbert's Hotel Problem. I think that the idea works fine. $\endgroup$ – Corrêa Mar 12 '19 at 4:08
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Try this:

We have that there exists a bijection $f:A \to \mathbb{N}$

Then $g = f^{-1}:\mathbb{N} \to A$ is a bijection too

And $B = \{b_{1},...,b_{k}\} \subset A$

So, we have that for every $n \in \mathbb{N}$, $g(n) = a_{n} \in A$

And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 \leq i \leq k$

And suppose that $n_{1}<n_{2}< \dots <n_{k}$

We are going to create a bijection $h:\mathbb{N} \to$ $A$ \ $B$

Let $d \in \mathbb{N}\cup\{0\}$, $d = n_{k} - k$

We have a subset with $d$ elements $\{1,2,...,n_{k}\}$\ $\{n_{1},...n_{k}\} = \{c_{1},...,c_{d}\}$

Let $h(n) = g(c_{n})$ if $1 \leq n \leq d$, and $h(n) = g(n + k)$ if $n>d$

Then $h$ define a bijection between $\mathbb{N}$ and $A$ \ $B$

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First, we observe that $A \setminus B$ is not finite, otherwise $A = A \setminus B \cup B$ would be finite. Now, $A \setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A \setminus B \subseteq A$ and $A$ is denumerable, $A \setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A \setminus B$ is denumerable.

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  • $\begingroup$ You mean $A$\ $B$ cannot be non-denumerable? $\endgroup$ – J. W. Tanner Mar 12 '19 at 4:11

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