Prove $|A times A| geq |A|$
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For case 1, I proved that if $A$ is finite/countable, then $|Atimes A|$ is finite/countable. However, I don’t know how to proceed with showing $|Atimes A|geq|A|$. Specifically, I want to say for every $a_i$ in $A$, there is a corresponding $(a_i, a_k)$ in $|Atimes A|$ whose list is certainly greater than a single $a_i$ for a corresponding $i$. However, I was given a remark which states if $|A| geq |B|$, then there exists some $f:A to B$ that is onto. How would I show $f:|A times A|to|A|$ is onto?
Case 2 would be for $A$ is not countable.
discrete-mathematics proof-writing
asked Nov 20 at 4:14
darylnak
156111
add a comment |
up vote
-2
down vote
favorite
For case 1, I proved that if $A$ is finite/countable, then $|Atimes A|$ is finite/countable. However, I don’t know how to proceed with showing $|Atimes A|geq|A|$. Specifically, I want to say for every $a_i$ in $A$, there is a corresponding $(a_i, a_k)$ in $|Atimes A|$ whose list is certainly greater than a single $a_i$ for a corresponding $i$. However, I was given a remark which states if $|A| geq |B|$, then there exists some $f:A to B$ that is onto. How would I show $f:|A times A|to|A|$ is onto?
Case 2 would be for $A$ is not countable.
discrete-mathematics proof-writing
asked Nov 20 at 4:14
darylnak
156111
Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
For case 1, I proved that if $A$ is finite/countable, then $|Atimes A|$ is finite/countable. However, I don’t know how to proceed with showing $|Atimes A|geq|A|$. Specifically, I want to say for every $a_i$ in $A$, there is a corresponding $(a_i, a_k)$ in $|Atimes A|$ whose list is certainly greater than a single $a_i$ for a corresponding $i$. However, I was given a remark which states if $|A| geq |B|$, then there exists some $f:A to B$ that is onto. How would I show $f:|A times A|to|A|$ is onto?
Case 2 would be for $A$ is not countable.
discrete-mathematics proof-writing
asked Nov 20 at 4:14
darylnak
156111
For case 1, I proved that if $A$ is finite/countable, then $|Atimes A|$ is finite/countable. However, I don’t know how to proceed with showing $|Atimes A|geq|A|$. Specifically, I want to say for every $a_i$ in $A$, there is a corresponding $(a_i, a_k)$ in $|Atimes A|$ whose list is certainly greater than a single $a_i$ for a corresponding $i$. However, I was given a remark which states if $|A| geq |B|$, then there exists some $f:A to B$ that is onto. How would I show $f:|A times A|to|A|$ is onto?
Case 2 would be for $A$ is not countable.
discrete-mathematics proof-writing
discrete-mathematics proof-writing
asked Nov 20 at 4:14
darylnak
156111
asked Nov 20 at 4:14
darylnak
156111
edited Nov 20 at 4:29
asked Nov 20 at 4:14
darylnak
156111
asked Nov 20 at 4:14
darylnak
156111
asked Nov 20 at 4:14
darylnak
156111
156111
Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21
add a comment |
Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21
Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = {1,2}$$
then
$$A times A = {(1,1);,;(1,2);,;(2,1);,;(2,2)}$$
We can define the function $f : Atimes A rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = emptyset$, which you can handle separately and more trivially.
answered Nov 20 at 4:22
Eevee Trainer
1,839216
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = {1,2}$$
then
$$A times A = {(1,1);,;(1,2);,;(2,1);,;(2,2)}$$
We can define the function $f : Atimes A rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = emptyset$, which you can handle separately and more trivially.
answered Nov 20 at 4:22
Eevee Trainer
1,839216
add a comment |
up vote
1
down vote
accepted
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = {1,2}$$
then
$$A times A = {(1,1);,;(1,2);,;(2,1);,;(2,2)}$$
We can define the function $f : Atimes A rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = emptyset$, which you can handle separately and more trivially.
answered Nov 20 at 4:22
Eevee Trainer
1,839216
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = {1,2}$$
then
$$A times A = {(1,1);,;(1,2);,;(2,1);,;(2,2)}$$
We can define the function $f : Atimes A rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = emptyset$, which you can handle separately and more trivially.
answered Nov 20 at 4:22
Eevee Trainer
1,839216
To show that $f$ is onto, consider a sort of projection. For example, if
$$A = {1,2}$$
then
$$A times A = {(1,1);,;(1,2);,;(2,1);,;(2,2)}$$
We can define the function $f : Atimes A rightarrow A $ by
$$f(x,y)=x$$
Then $(1,1)$ and $(1,2)$ map to $1$, and $(2,1)$ and $(2,2)$ map to $2$. Thus, every element of $A$ has a preimage.
Can you see how this might prove $f$ is surjective in a more general case if $A$ is nonempty? (Or "onto," whichever terminology you prefer.) About the only nuance you would really need to consider for this would be $A = emptyset$, which you can handle separately and more trivially.
answered Nov 20 at 4:22
Eevee Trainer
1,839216
answered Nov 20 at 4:22
Eevee Trainer
1,839216
answered Nov 20 at 4:22
Eevee Trainer
1,839216
answered Nov 20 at 4:22
Eevee Trainer
1,839216
1,839216
add a comment |
add a comment |
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Please use the MathJax standard format and also wrtie down the problem and what you are doing or trying. As is, the problem will be closed shortly.
– Will M.
Nov 20 at 4:16
Hint: Assume $A$ is nonempty. Consider a specific $xin A$. Consider the set ${(a,x)~:~ain A}$ and how it compares to $A$ itself. Additional hint: $|Y|geq |X|$ if and only if there is some $Zsubseteq Y$ such that $|Z|=|X|$.
– JMoravitz
Nov 20 at 4:21