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Can I say “let $f : mathbbR mapsto mathbbR$†without the axiom of choice? [duplicate]
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Do we need Axiom of Choice to make infinite choices from a set?
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Indeed letting $ f$ be in $mathbbR^mathbbR$ seems like it requires making $2^aleph_0$ choices.
axiom-of-choice
asked Jul 19 at 22:55
Raphael Picovschi
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marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Jul 20 at 0:11
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This question already has an answer here:
Do we need Axiom of Choice to make infinite choices from a set?
5 answers
Indeed letting $ f$ be in $mathbbR^mathbbR$ seems like it requires making $2^aleph_0$ choices.
axiom-of-choice
asked Jul 19 at 22:55
Raphael Picovschi
235
marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Jul 20 at 0:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06
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up vote
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down vote
favorite
This question already has an answer here:
Do we need Axiom of Choice to make infinite choices from a set?
5 answers
Indeed letting $ f$ be in $mathbbR^mathbbR$ seems like it requires making $2^aleph_0$ choices.
axiom-of-choice
asked Jul 19 at 22:55
Raphael Picovschi
235
This question already has an answer here:
Do we need Axiom of Choice to make infinite choices from a set?
5 answers
Indeed letting $ f$ be in $mathbbR^mathbbR$ seems like it requires making $2^aleph_0$ choices.
This question already has an answer here:
Do we need Axiom of Choice to make infinite choices from a set?
5 answers
axiom-of-choice
asked Jul 19 at 22:55
Raphael Picovschi
235
asked Jul 19 at 22:55
Raphael Picovschi
235
asked Jul 19 at 22:55
Raphael Picovschi
235
asked Jul 19 at 22:55
Raphael Picovschi
235
235
marked as duplicate by Asaf Karagila♦ axiom-of-choice
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Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06
add a comment |Â
1
Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06
1
1
Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06
Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06
add a comment |Â
1 Answer
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The set of functions with domain $mathbb R$ and range $mathbb R$ is a set. That set is nonempty. So, like any other set, you are allowed to pick one element out of that set, and to do it with a phrase like "Let $f$ be an element of that set".
What the axiom of choice asserts is the nonemptiness of certain function sets, under certain constraints. For example, suppose you already had a function $A : mathbb R to P(mathbb R)$, and so for each $r in mathbb R$ you have a nomempty subset $A(r) subset mathbb R$. Consider the set of all functions $f : mathbb R to mathbb R$ such that $f(r) in A(r)$ for every $r in mathbb R$. The axiom of choice, as applied to this situation, asserts that this set of functions is not empty, i.e. that there exists a function $f$ satisfying those conditions.
answered Jul 19 at 23:08
Lee Mosher
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1 Answer
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up vote
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The set of functions with domain $mathbb R$ and range $mathbb R$ is a set. That set is nonempty. So, like any other set, you are allowed to pick one element out of that set, and to do it with a phrase like "Let $f$ be an element of that set".
What the axiom of choice asserts is the nonemptiness of certain function sets, under certain constraints. For example, suppose you already had a function $A : mathbb R to P(mathbb R)$, and so for each $r in mathbb R$ you have a nomempty subset $A(r) subset mathbb R$. Consider the set of all functions $f : mathbb R to mathbb R$ such that $f(r) in A(r)$ for every $r in mathbb R$. The axiom of choice, as applied to this situation, asserts that this set of functions is not empty, i.e. that there exists a function $f$ satisfying those conditions.
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
add a comment |Â
up vote
5
down vote
The set of functions with domain $mathbb R$ and range $mathbb R$ is a set. That set is nonempty. So, like any other set, you are allowed to pick one element out of that set, and to do it with a phrase like "Let $f$ be an element of that set".
What the axiom of choice asserts is the nonemptiness of certain function sets, under certain constraints. For example, suppose you already had a function $A : mathbb R to P(mathbb R)$, and so for each $r in mathbb R$ you have a nomempty subset $A(r) subset mathbb R$. Consider the set of all functions $f : mathbb R to mathbb R$ such that $f(r) in A(r)$ for every $r in mathbb R$. The axiom of choice, as applied to this situation, asserts that this set of functions is not empty, i.e. that there exists a function $f$ satisfying those conditions.
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The set of functions with domain $mathbb R$ and range $mathbb R$ is a set. That set is nonempty. So, like any other set, you are allowed to pick one element out of that set, and to do it with a phrase like "Let $f$ be an element of that set".
What the axiom of choice asserts is the nonemptiness of certain function sets, under certain constraints. For example, suppose you already had a function $A : mathbb R to P(mathbb R)$, and so for each $r in mathbb R$ you have a nomempty subset $A(r) subset mathbb R$. Consider the set of all functions $f : mathbb R to mathbb R$ such that $f(r) in A(r)$ for every $r in mathbb R$. The axiom of choice, as applied to this situation, asserts that this set of functions is not empty, i.e. that there exists a function $f$ satisfying those conditions.
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
The set of functions with domain $mathbb R$ and range $mathbb R$ is a set. That set is nonempty. So, like any other set, you are allowed to pick one element out of that set, and to do it with a phrase like "Let $f$ be an element of that set".
What the axiom of choice asserts is the nonemptiness of certain function sets, under certain constraints. For example, suppose you already had a function $A : mathbb R to P(mathbb R)$, and so for each $r in mathbb R$ you have a nomempty subset $A(r) subset mathbb R$. Consider the set of all functions $f : mathbb R to mathbb R$ such that $f(r) in A(r)$ for every $r in mathbb R$. The axiom of choice, as applied to this situation, asserts that this set of functions is not empty, i.e. that there exists a function $f$ satisfying those conditions.
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
edited Jul 19 at 23:13
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
answered Jul 19 at 23:08
Lee Mosher
45.6k33478
45.6k33478
add a comment |Â
add a comment |Â
1
Your "let ..." is not mathematical proposition, it is just preparing for some proposition of form e.g. "every function $f : mathbbR to mathbbR$ has property P", you can do such propositions without AC
– kp9r4d
Jul 19 at 23:06