Let A be a denumerable subset of an uncountable set X. Prove that X/A is uncountable
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I'd like to know if my proof is correct:
Assume $A$ is denumerable and and $Asubseteq X$
Assume X is uncountable
(X uncountable means X is not finite and X is not denumerable)
$X=(Xbackslash A) cup (X cap A)=(Xbackslash A)cup A$
(as $A subseteq X$)
Therefore $X$ is uncountable $implies (Xbackslash A) cup (X cap A)$ uncountable.
Now using the Theorem: If A and B are denumerable sets then $Acup B$ is denumerable.
The contrapositive of this statement is: If $Acup B$ is not denumerable then A is not denumerable or B is not denumerable.
Now applying the contrapositive to what we have so far
$(Xbackslash A) cup A)$ uncountable(not denumerable) $implies(Xbackslash A)$ uncountable or $A$ is uncountable.
(by the contrapositive of the theorem)
By assumption A is denumerable and therefore countable hence XA is uncountable as required.
proof-verification elementary-set-theory
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
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add a comment |
$begingroup$
I'd like to know if my proof is correct:
Assume $A$ is denumerable and and $Asubseteq X$
Assume X is uncountable
(X uncountable means X is not finite and X is not denumerable)
$X=(Xbackslash A) cup (X cap A)=(Xbackslash A)cup A$
(as $A subseteq X$)
Therefore $X$ is uncountable $implies (Xbackslash A) cup (X cap A)$ uncountable.
Now using the Theorem: If A and B are denumerable sets then $Acup B$ is denumerable.
The contrapositive of this statement is: If $Acup B$ is not denumerable then A is not denumerable or B is not denumerable.
Now applying the contrapositive to what we have so far
$(Xbackslash A) cup A)$ uncountable(not denumerable) $implies(Xbackslash A)$ uncountable or $A$ is uncountable.
(by the contrapositive of the theorem)
By assumption A is denumerable and therefore countable hence XA is uncountable as required.
proof-verification elementary-set-theory
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
$endgroup$
$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
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– Thomas Shelby
Jan 8 at 11:22
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Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36
add a comment |
$begingroup$
I'd like to know if my proof is correct:
Assume $A$ is denumerable and and $Asubseteq X$
Assume X is uncountable
(X uncountable means X is not finite and X is not denumerable)
$X=(Xbackslash A) cup (X cap A)=(Xbackslash A)cup A$
(as $A subseteq X$)
Therefore $X$ is uncountable $implies (Xbackslash A) cup (X cap A)$ uncountable.
Now using the Theorem: If A and B are denumerable sets then $Acup B$ is denumerable.
The contrapositive of this statement is: If $Acup B$ is not denumerable then A is not denumerable or B is not denumerable.
Now applying the contrapositive to what we have so far
$(Xbackslash A) cup A)$ uncountable(not denumerable) $implies(Xbackslash A)$ uncountable or $A$ is uncountable.
(by the contrapositive of the theorem)
By assumption A is denumerable and therefore countable hence XA is uncountable as required.
proof-verification elementary-set-theory
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
$endgroup$
I'd like to know if my proof is correct:
Assume $A$ is denumerable and and $Asubseteq X$
Assume X is uncountable
(X uncountable means X is not finite and X is not denumerable)
$X=(Xbackslash A) cup (X cap A)=(Xbackslash A)cup A$
(as $A subseteq X$)
Therefore $X$ is uncountable $implies (Xbackslash A) cup (X cap A)$ uncountable.
Now using the Theorem: If A and B are denumerable sets then $Acup B$ is denumerable.
The contrapositive of this statement is: If $Acup B$ is not denumerable then A is not denumerable or B is not denumerable.
Now applying the contrapositive to what we have so far
$(Xbackslash A) cup A)$ uncountable(not denumerable) $implies(Xbackslash A)$ uncountable or $A$ is uncountable.
(by the contrapositive of the theorem)
By assumption A is denumerable and therefore countable hence XA is uncountable as required.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
asked Jan 8 at 11:16
stochasticmrfoxstochasticmrfox
837
837
$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
$endgroup$
– Thomas Shelby
Jan 8 at 11:22
$begingroup$
Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36
add a comment |
$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
$endgroup$
– Thomas Shelby
Jan 8 at 11:22
$begingroup$
Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36
$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
$endgroup$
– Thomas Shelby
Jan 8 at 11:22
$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
$endgroup$
– Thomas Shelby
Jan 8 at 11:22
$begingroup$
Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36
$begingroup$
Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36
add a comment |
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$begingroup$
Your proof seems fine, but it's "better" to start with assuming $Xsetminus A$ is denumerable and arrive at a contradiction.
$endgroup$
– Thomas Shelby
Jan 8 at 11:22
$begingroup$
Yes, it might be better to say something like: Suppose $Xsetminus A$ is denumerable, then $(Xsetminus A)cup A=X$ is denumerable, contradiction."
$endgroup$
– palmpo
Jan 8 at 15:36