Prove the existence of a one-to-one function for a recursive definition
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Specifically, I don’t know what $g(x) = {f(x)}$ means, thus I’m not sure how to proceed with the proof. Is it implied that I need to define another function $f(x)$ or do I need to prove the latter $f(x)$?
discrete-mathematics proof-writing proof-explanation
asked 8 hours ago
darylnak
140111
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up vote
0
down vote
favorite
Specifically, I don’t know what $g(x) = {f(x)}$ means, thus I’m not sure how to proceed with the proof. Is it implied that I need to define another function $f(x)$ or do I need to prove the latter $f(x)$?
discrete-mathematics proof-writing proof-explanation
asked 8 hours ago
darylnak
140111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Specifically, I don’t know what $g(x) = {f(x)}$ means, thus I’m not sure how to proceed with the proof. Is it implied that I need to define another function $f(x)$ or do I need to prove the latter $f(x)$?
discrete-mathematics proof-writing proof-explanation
asked 8 hours ago
darylnak
140111
Specifically, I don’t know what $g(x) = {f(x)}$ means, thus I’m not sure how to proceed with the proof. Is it implied that I need to define another function $f(x)$ or do I need to prove the latter $f(x)$?
discrete-mathematics proof-writing proof-explanation
discrete-mathematics proof-writing proof-explanation
asked 8 hours ago
darylnak
140111
asked 8 hours ago
darylnak
140111
asked 8 hours ago
darylnak
140111
asked 8 hours ago
darylnak
140111
asked 8 hours ago
darylnak
140111
140111
add a comment |
add a comment |
1 Answer
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Now you want to define a function $g:Atomathcal{P}(Y)$, that is, for any element $xin A$, you need to associate a subset of $Y$. The subset is chosen to be the singleton ${f(x)}$.
It is of course a well-defined function. The problem is to show that $g$ is one-to-one. However, since $f$ is one-to-one, we have $g(x)=g(x')implies {f(x)}={f(x')}implies f(x)=f(x')implies x=x'$.
answered 8 hours ago
Eclipse Sun
6,7721337
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Now you want to define a function $g:Atomathcal{P}(Y)$, that is, for any element $xin A$, you need to associate a subset of $Y$. The subset is chosen to be the singleton ${f(x)}$.
It is of course a well-defined function. The problem is to show that $g$ is one-to-one. However, since $f$ is one-to-one, we have $g(x)=g(x')implies {f(x)}={f(x')}implies f(x)=f(x')implies x=x'$.
answered 8 hours ago
Eclipse Sun
6,7721337
add a comment |
up vote
1
down vote
accepted
Now you want to define a function $g:Atomathcal{P}(Y)$, that is, for any element $xin A$, you need to associate a subset of $Y$. The subset is chosen to be the singleton ${f(x)}$.
It is of course a well-defined function. The problem is to show that $g$ is one-to-one. However, since $f$ is one-to-one, we have $g(x)=g(x')implies {f(x)}={f(x')}implies f(x)=f(x')implies x=x'$.
answered 8 hours ago
Eclipse Sun
6,7721337
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Now you want to define a function $g:Atomathcal{P}(Y)$, that is, for any element $xin A$, you need to associate a subset of $Y$. The subset is chosen to be the singleton ${f(x)}$.
It is of course a well-defined function. The problem is to show that $g$ is one-to-one. However, since $f$ is one-to-one, we have $g(x)=g(x')implies {f(x)}={f(x')}implies f(x)=f(x')implies x=x'$.
answered 8 hours ago
Eclipse Sun
6,7721337
Now you want to define a function $g:Atomathcal{P}(Y)$, that is, for any element $xin A$, you need to associate a subset of $Y$. The subset is chosen to be the singleton ${f(x)}$.
It is of course a well-defined function. The problem is to show that $g$ is one-to-one. However, since $f$ is one-to-one, we have $g(x)=g(x')implies {f(x)}={f(x')}implies f(x)=f(x')implies x=x'$.
answered 8 hours ago
Eclipse Sun
6,7721337
answered 8 hours ago
Eclipse Sun
6,7721337
answered 8 hours ago
Eclipse Sun
6,7721337
answered 8 hours ago
Eclipse Sun
6,7721337
6,7721337
add a comment |
add a comment |
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