What is the cardinality of the set of all functions mapping ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$ into the...












0














Let $T = {sqrt{2},sqrt{3},sqrt{5},sqrt{7}}$ and $mathcal{S}$ the set of all functions that maps $T$ into $mathbb{Q}.$ What is the cardinality of $mathcal{S}$?



So far I've been messing with the idea that we can take the subset $T_0subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(sqrt{2})=(sqrt{2})^2+a$ where $a in mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=aleph_0.$



I'm not really sure if this is going to get me anywhere.






elementary-set-theory cardinals







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  • 3




    It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
    – Seewoo Lee
    Nov 24 at 5:15






  • 1




    The particular values of your set are not important. Only that it has four elements.
    – badjohn
    Nov 24 at 6:27


















0














Let $T = {sqrt{2},sqrt{3},sqrt{5},sqrt{7}}$ and $mathcal{S}$ the set of all functions that maps $T$ into $mathbb{Q}.$ What is the cardinality of $mathcal{S}$?



So far I've been messing with the idea that we can take the subset $T_0subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(sqrt{2})=(sqrt{2})^2+a$ where $a in mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=aleph_0.$



I'm not really sure if this is going to get me anywhere.






elementary-set-theory cardinals







share|cite|improve this question


















  • 3




    It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
    – Seewoo Lee
    Nov 24 at 5:15






  • 1




    The particular values of your set are not important. Only that it has four elements.
    – badjohn
    Nov 24 at 6:27
















0












0








0







Let $T = {sqrt{2},sqrt{3},sqrt{5},sqrt{7}}$ and $mathcal{S}$ the set of all functions that maps $T$ into $mathbb{Q}.$ What is the cardinality of $mathcal{S}$?



So far I've been messing with the idea that we can take the subset $T_0subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(sqrt{2})=(sqrt{2})^2+a$ where $a in mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=aleph_0.$



I'm not really sure if this is going to get me anywhere.






elementary-set-theory cardinals







share|cite|improve this question













Let $T = {sqrt{2},sqrt{3},sqrt{5},sqrt{7}}$ and $mathcal{S}$ the set of all functions that maps $T$ into $mathbb{Q}.$ What is the cardinality of $mathcal{S}$?



So far I've been messing with the idea that we can take the subset $T_0subset T$ of functions that square an element of $T$ and adds any other string of rational numbers to it. For example $f(sqrt{2})=(sqrt{2})^2+a$ where $a in mathbb{Q}.$ Since $|T_0|$ is countable and infinite $|T_0|=aleph_0.$



I'm not really sure if this is going to get me anywhere.






elementary-set-theory cardinals




elementary-set-theory cardinals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 5:12









Trevor Mason

197




197








  • 3




    It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
    – Seewoo Lee
    Nov 24 at 5:15






  • 1




    The particular values of your set are not important. Only that it has four elements.
    – badjohn
    Nov 24 at 6:27
















  • 3




    It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
    – Seewoo Lee
    Nov 24 at 5:15






  • 1




    The particular values of your set are not important. Only that it has four elements.
    – badjohn
    Nov 24 at 6:27










3




3




It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
– Seewoo Lee
Nov 24 at 5:15




It is same as the cardinality of $mathbb{Q}^{4}$, which is countable.
– Seewoo Lee
Nov 24 at 5:15




1




1




The particular values of your set are not important. Only that it has four elements.
– badjohn
Nov 24 at 6:27






The particular values of your set are not important. Only that it has four elements.
– badjohn
Nov 24 at 6:27












1 Answer
1






active

oldest

votes


















1














Could you answer the question of the cardinality of the set of all functions from ${ 1 }$ to $mathbb{Q}$? I hope so, it's obviously the same as $mathbb{Q}$ since any such function can be specified by a single value from $mathbb{Q}$.



Now, how about the cardinality of the set of all functions from ${ 1, 2 }$ to $mathbb{Q}$? Well, any such function can be specified by two values from $mathbb{Q}$ so its cardinality is the same as $mathbb{Q} times mathbb{Q} = mathbb{Q}^2$. Do you know the cardinality of this set?



Let's jump a step to the set of functions from ${1, 2, 3, 4 }$ to $mathbb{Q}$. As Seewoo says, this has cardinality $mathbb{Q} times mathbb{Q} times mathbb{Q} times mathbb{Q} = mathbb{Q}^4$. If you figured that $mathbb{Q}^2$ is countable then you should see that this is as well.



Now your set: ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$. Well there is a very simple bijection between it and my set ${1, 2, 3, 4 }$ which can be easily used to associate the maps from my set to $mathbb{Q}$ with those from your set to $mathbb{Q}$.






share|cite|improve this answer





















  • I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
    – Trevor Mason
    Nov 25 at 0:36










  • @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
    – badjohn
    Nov 25 at 8:42












  • This should have cardinality $mathbb{Q}^5?$
    – Trevor Mason
    Nov 25 at 21:39










  • Yes, but what is that?
    – badjohn
    Nov 25 at 22:12










  • $mathbb{Q}= aleph_0?$
    – Trevor Mason
    Nov 25 at 22:13











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1 Answer
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1 Answer
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active

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active

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1














Could you answer the question of the cardinality of the set of all functions from ${ 1 }$ to $mathbb{Q}$? I hope so, it's obviously the same as $mathbb{Q}$ since any such function can be specified by a single value from $mathbb{Q}$.



Now, how about the cardinality of the set of all functions from ${ 1, 2 }$ to $mathbb{Q}$? Well, any such function can be specified by two values from $mathbb{Q}$ so its cardinality is the same as $mathbb{Q} times mathbb{Q} = mathbb{Q}^2$. Do you know the cardinality of this set?



Let's jump a step to the set of functions from ${1, 2, 3, 4 }$ to $mathbb{Q}$. As Seewoo says, this has cardinality $mathbb{Q} times mathbb{Q} times mathbb{Q} times mathbb{Q} = mathbb{Q}^4$. If you figured that $mathbb{Q}^2$ is countable then you should see that this is as well.



Now your set: ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$. Well there is a very simple bijection between it and my set ${1, 2, 3, 4 }$ which can be easily used to associate the maps from my set to $mathbb{Q}$ with those from your set to $mathbb{Q}$.






share|cite|improve this answer





















  • I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
    – Trevor Mason
    Nov 25 at 0:36










  • @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
    – badjohn
    Nov 25 at 8:42












  • This should have cardinality $mathbb{Q}^5?$
    – Trevor Mason
    Nov 25 at 21:39










  • Yes, but what is that?
    – badjohn
    Nov 25 at 22:12










  • $mathbb{Q}= aleph_0?$
    – Trevor Mason
    Nov 25 at 22:13
















1














Could you answer the question of the cardinality of the set of all functions from ${ 1 }$ to $mathbb{Q}$? I hope so, it's obviously the same as $mathbb{Q}$ since any such function can be specified by a single value from $mathbb{Q}$.



Now, how about the cardinality of the set of all functions from ${ 1, 2 }$ to $mathbb{Q}$? Well, any such function can be specified by two values from $mathbb{Q}$ so its cardinality is the same as $mathbb{Q} times mathbb{Q} = mathbb{Q}^2$. Do you know the cardinality of this set?



Let's jump a step to the set of functions from ${1, 2, 3, 4 }$ to $mathbb{Q}$. As Seewoo says, this has cardinality $mathbb{Q} times mathbb{Q} times mathbb{Q} times mathbb{Q} = mathbb{Q}^4$. If you figured that $mathbb{Q}^2$ is countable then you should see that this is as well.



Now your set: ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$. Well there is a very simple bijection between it and my set ${1, 2, 3, 4 }$ which can be easily used to associate the maps from my set to $mathbb{Q}$ with those from your set to $mathbb{Q}$.






share|cite|improve this answer





















  • I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
    – Trevor Mason
    Nov 25 at 0:36










  • @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
    – badjohn
    Nov 25 at 8:42












  • This should have cardinality $mathbb{Q}^5?$
    – Trevor Mason
    Nov 25 at 21:39










  • Yes, but what is that?
    – badjohn
    Nov 25 at 22:12










  • $mathbb{Q}= aleph_0?$
    – Trevor Mason
    Nov 25 at 22:13














1












1








1






Could you answer the question of the cardinality of the set of all functions from ${ 1 }$ to $mathbb{Q}$? I hope so, it's obviously the same as $mathbb{Q}$ since any such function can be specified by a single value from $mathbb{Q}$.



Now, how about the cardinality of the set of all functions from ${ 1, 2 }$ to $mathbb{Q}$? Well, any such function can be specified by two values from $mathbb{Q}$ so its cardinality is the same as $mathbb{Q} times mathbb{Q} = mathbb{Q}^2$. Do you know the cardinality of this set?



Let's jump a step to the set of functions from ${1, 2, 3, 4 }$ to $mathbb{Q}$. As Seewoo says, this has cardinality $mathbb{Q} times mathbb{Q} times mathbb{Q} times mathbb{Q} = mathbb{Q}^4$. If you figured that $mathbb{Q}^2$ is countable then you should see that this is as well.



Now your set: ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$. Well there is a very simple bijection between it and my set ${1, 2, 3, 4 }$ which can be easily used to associate the maps from my set to $mathbb{Q}$ with those from your set to $mathbb{Q}$.






share|cite|improve this answer












Could you answer the question of the cardinality of the set of all functions from ${ 1 }$ to $mathbb{Q}$? I hope so, it's obviously the same as $mathbb{Q}$ since any such function can be specified by a single value from $mathbb{Q}$.



Now, how about the cardinality of the set of all functions from ${ 1, 2 }$ to $mathbb{Q}$? Well, any such function can be specified by two values from $mathbb{Q}$ so its cardinality is the same as $mathbb{Q} times mathbb{Q} = mathbb{Q}^2$. Do you know the cardinality of this set?



Let's jump a step to the set of functions from ${1, 2, 3, 4 }$ to $mathbb{Q}$. As Seewoo says, this has cardinality $mathbb{Q} times mathbb{Q} times mathbb{Q} times mathbb{Q} = mathbb{Q}^4$. If you figured that $mathbb{Q}^2$ is countable then you should see that this is as well.



Now your set: ${sqrt{2},sqrt{3}, sqrt{5}, sqrt{7}}$. Well there is a very simple bijection between it and my set ${1, 2, 3, 4 }$ which can be easily used to associate the maps from my set to $mathbb{Q}$ with those from your set to $mathbb{Q}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 12:12









badjohn

4,2221620




4,2221620












  • I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
    – Trevor Mason
    Nov 25 at 0:36










  • @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
    – badjohn
    Nov 25 at 8:42












  • This should have cardinality $mathbb{Q}^5?$
    – Trevor Mason
    Nov 25 at 21:39










  • Yes, but what is that?
    – badjohn
    Nov 25 at 22:12










  • $mathbb{Q}= aleph_0?$
    – Trevor Mason
    Nov 25 at 22:13


















  • I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
    – Trevor Mason
    Nov 25 at 0:36










  • @TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
    – badjohn
    Nov 25 at 8:42












  • This should have cardinality $mathbb{Q}^5?$
    – Trevor Mason
    Nov 25 at 21:39










  • Yes, but what is that?
    – badjohn
    Nov 25 at 22:12










  • $mathbb{Q}= aleph_0?$
    – Trevor Mason
    Nov 25 at 22:13
















I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
– Trevor Mason
Nov 25 at 0:36




I absolutely understand this. Thank you so much. I got so caught up in the values of the set and not how many elements are in the set. This is a new level of abstraction for me compared to what I've been exposed to so far at the undergraduate level.
– Trevor Mason
Nov 25 at 0:36












@TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
– badjohn
Nov 25 at 8:42






@TrevorMason. I expect that was the intention of whoever wrote the question. How about the set ${ 0, 1, e, i, pi }$?
– badjohn
Nov 25 at 8:42














This should have cardinality $mathbb{Q}^5?$
– Trevor Mason
Nov 25 at 21:39




This should have cardinality $mathbb{Q}^5?$
– Trevor Mason
Nov 25 at 21:39












Yes, but what is that?
– badjohn
Nov 25 at 22:12




Yes, but what is that?
– badjohn
Nov 25 at 22:12












$mathbb{Q}= aleph_0?$
– Trevor Mason
Nov 25 at 22:13




$mathbb{Q}= aleph_0?$
– Trevor Mason
Nov 25 at 22:13


















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