Demonstrate $A subseteq P(cup A)$
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
$A$ is a set
$P(A)=B$
$ cup A = x $
My try:
If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$
then $xsubseteq P(cup A)$
I fell like this last step is not quite right.
elementary-set-theory
asked Sep 10 at 18:51
Sebas Martinez Santos
106
add a comment |Â
up vote
-2
down vote
favorite
$A$ is a set
$P(A)=B$
$ cup A = x $
My try:
If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$
then $xsubseteq P(cup A)$
I fell like this last step is not quite right.
elementary-set-theory
asked Sep 10 at 18:51
Sebas Martinez Santos
106
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
$A$ is a set
$P(A)=B$
$ cup A = x $
My try:
If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$
then $xsubseteq P(cup A)$
I fell like this last step is not quite right.
elementary-set-theory
asked Sep 10 at 18:51
Sebas Martinez Santos
106
$A$ is a set
$P(A)=B$
$ cup A = x $
My try:
If $Csubseteq A$ if there is $xin C$, tal que $xin cup A$
then $xsubseteq P(cup A)$
I fell like this last step is not quite right.
elementary-set-theory
elementary-set-theory
asked Sep 10 at 18:51
Sebas Martinez Santos
106
asked Sep 10 at 18:51
Sebas Martinez Santos
106
edited Sep 10 at 19:06
asked Sep 10 at 18:51
Sebas Martinez Santos
106
asked Sep 10 at 18:51
Sebas Martinez Santos
106
asked Sep 10 at 18:51
Sebas Martinez Santos
106
106
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$
Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$
answered Sep 10 at 18:59
Riccardo Ceccon
875320
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$
Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$
answered Sep 10 at 18:59
Riccardo Ceccon
875320
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
add a comment |Â
up vote
1
down vote
Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$
Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$
answered Sep 10 at 18:59
Riccardo Ceccon
875320
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$
Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$
answered Sep 10 at 18:59
Riccardo Ceccon
875320
Take $xin A:$ you want to prove that it is an element of $mathcalP(cup A),$ so you prove that $Asubseteq mathcalP(cup A).$
Actually you can write $x$ as $ ain x.$ From this point of view it is clear that $x$ is a subset of $cup A$ and so it belongs to the set $mathcalP(cup A).$
answered Sep 10 at 18:59
Riccardo Ceccon
875320
answered Sep 10 at 18:59
Riccardo Ceccon
875320
answered Sep 10 at 18:59
Riccardo Ceccon
875320
answered Sep 10 at 18:59
Riccardo Ceccon
875320
875320
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
add a comment |Â
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
But by definition $cup A$ does not have any subsets, they are just elements. For example: $A=1,2,3$ then $cup A=1,2,3$. What do you think? }
– Sebas Martinez Santos
Sep 10 at 19:17
2
2
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
@SebasMartinezSantos You are correct that $bigcup A = 1,2,3$ in your example. You say "by definition $bigcup A$ does not have any subsets..." Every set has subsets, including those sets which are sets of sets or otherwise. In your example $bigcup A$ has the following eight subsets: $emptyset, 1,2,3,1,2dots,1,2,3$. As such the power set of $bigcup A$ would be $emptyset,1,2,dots,2,3,1,2,3$. You may have meant to say $bigcup A$ does not have any sets as elements, which has a different meaning.
– JMoravitz
Sep 10 at 19:33
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
Exactly, be careful Sebas that an element of A can be a set itself.
– Riccardo Ceccon
Sep 10 at 20:10
add a comment |Â
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