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If $B$ is an uncountable set and $A$ is a countable set, then prove that $B$ is similar to $B-A$.

Attempt:

Two sets $A$ and $B$ are called similar $\iff$ thee exists a one to one function $F$ whose domain is the set $A$ and whose range is the set $B$.

$B$ is an uncountable set and $A$ is a countable set, then $B-A$ must also be uncountable and hence, an infinite set.

To prove this, let us suppose $B-A$ is countable. Since, $A$ is countable, hence, $(B-A) \cup A$ should also be countable. But $B \subseteq (B-A) \cup A$ should be countable as well which is a contradiction.

Hence, $B-A$ is an uncountable set as well. Now, we need to show that $(B-A) \sim B$.

How do I move ahead? I think we need to define a one-one onto function from $(B-A) \rightarrow B$ but I am not able to think of such a function.

Thank you for your help.

MathMan
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    Hint: Use the same idea as in Hilbert's hotel. To make room for countably many new guests, take countably many rooms $1,2,3,\ldots$ and move the guests in those rooms to rooms $2,4,6,\ldots$. – Samuel Sep 27 '14 at 11:10

1 Answers1

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You showed that $B-A$ is uncountable, so you can extract another countable set from it.

Let's say that $B=A\cup (B-A)=A\cup C \cup B' $, where $C$ is countable and $B'$ uncountable, with $C$ and $B'$ disjoint.

It's easy to construct a bijection between $C$ and $A\cup C$, since are both countable sets, so you have a bijection from $C \cup B'=B-A$ and $A\cup C \cup B'=B$

This works if $A\subseteq B$. If this isn't true, then let $A'$ be $A\cap B$, and repeat the same reasoning with $A'$, and it works since $B-A=B-A'$

Exodd
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  • Isn't $A \cup (B-A) = A \cup B ?$ – MathMan Sep 27 '14 at 11:20
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    If $A$ isn't included in $B$, sobstitute $A'=A\cap B$ to $A$, and the same reasoning works – Exodd Sep 27 '14 at 11:22
  • okay .. I understood till the point that there is a bijection between $C$ and $A \cup C$. But, how does this ensure bijection between $C \cup B'$ and $A \cup C \cup B'$? – MathMan Sep 27 '14 at 11:35
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    $A$, $C$ and $B'$ are disjoint sets, so construct the function that is the identity from $B'$ to $B'$, and goes bijectively from $C$ to $A\cup C$ – Exodd Sep 27 '14 at 11:44
  • $B'$ is an uncountable set but $C$ and $A \cup C$ are countable sets. How can we sure that such a bijection exists? Their cardinalities are different? – MathMan Sep 27 '14 at 11:49
  • $C$ is an infinite countable set, that is in bijective relation with $\mathbb{N}$, and the same holds for $A\cup C$ – Exodd Sep 27 '14 at 11:54
  • yeah .. I was talking about the bijection between set $B' \rightarrow C$ – MathMan Sep 27 '14 at 11:58
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    why do you want that bijection? Follow me. You want a bijection between $B-A$ and $B$. If $A\subseteq B$, then $B-A=C\cup B'$ and $B=A\cup C\cup B'$, where $A$, $C$ and $B'$ are disjoint sets. So you construct a bijection $B\to B-A$ that goes from $B'$ to $B'$ and from $A\cup C$ to $C$. – Exodd Sep 27 '14 at 12:03
  • Oh okay .. Got it.. Thank you very much. – MathMan Sep 27 '14 at 12:19