Can a relation that is NOT a function be one-to-one or onto? The Next CEO of Stack OverflowHow...
It is correct to match light sources with the same color temperature?
Is French Guiana a (hard) EU border?
The Ultimate Number Sequence Puzzle
TikZ: How to fill area with a special pattern?
What flight has the highest ratio of timezone difference to flight time?
Is there a reasonable and studied concept of reduction between regular languages?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Is Nisuin Biblical or Rabbinic?
Do scriptures give a method to recognize a truly self-realized person/jivanmukta?
how one can write a nice vector parser, something that does pgfvecparse{A=B-C; D=E x F;}
How to Implement Deterministic Encryption Safely in .NET
What are the unusually-enlarged wing sections on this P-38 Lightning?
(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?
Reshaping json / reparing json inside shell script (remove trailing comma)
How to avoid supervisors with prejudiced views?
Is it okay to majorly distort historical facts while writing a fiction story?
Purpose of level-shifter with same in and out voltages
Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens
Vector calculus integration identity problem
How do I fit a non linear curve?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
Calculate the Mean mean of two numbers
What was Carter Burke's job for "the company" in Aliens?
What does "shotgun unity" refer to here in this sentence?
Can a relation that is NOT a function be one-to-one or onto?
The Next CEO of Stack OverflowHow do we make the domain of a inverse function if the function is not ontoDetermining If A Relation Is A FunctionIs this function one-to-one, onto, both or not a function?Given 2 sets (X and Y) is it possible for $f: Y to X $ to be a relation, or not?Finding if a function is onto?Proving that a relation is an equivalence relationProof of one-to-one and onto for a functionOne to One and Onto functionsOne-to-One and Onto ProofDifficulty understanding this example that proves a function is one-to-one
$begingroup$
Let $A = {1,2,3}$ and $B = {7,8,9}$. For the relation between $A$ and $B$ given as a subset of $A times B = {(1,9), (1,7), (3,8)}$, $Atimes B$ is not a function because $1$ appears as the first element in more than one ordered pair. This made me think about the following generalized question.
If a relation is not a function, then it is only a subset of $A$ and $B$ denoted as $Atimes B$. Then can this subset be one-to-one? Can it be onto?
One-to-One?: Even though $Atimes B$ is not a function, I believe it satisfies the requirements of being one-to-one because in $A$, $x_1 = x_2$ only when $x_1 = x_1$.
Onto?: I want to say yes also. The range of $A$ could be the range of $B$. If $A times B = {(1,9), (1,8), (3,8)}$, then does this satisfy the requirement of being onto?
Basically what is piquing my curiosity is if non-function relations can be one-to-one or onto or if being one-to-one and/or onto implies that it must be a function.
abstract-algebra functions
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
$endgroup$
add a comment |
$begingroup$
Let $A = {1,2,3}$ and $B = {7,8,9}$. For the relation between $A$ and $B$ given as a subset of $A times B = {(1,9), (1,7), (3,8)}$, $Atimes B$ is not a function because $1$ appears as the first element in more than one ordered pair. This made me think about the following generalized question.
If a relation is not a function, then it is only a subset of $A$ and $B$ denoted as $Atimes B$. Then can this subset be one-to-one? Can it be onto?
One-to-One?: Even though $Atimes B$ is not a function, I believe it satisfies the requirements of being one-to-one because in $A$, $x_1 = x_2$ only when $x_1 = x_1$.
Onto?: I want to say yes also. The range of $A$ could be the range of $B$. If $A times B = {(1,9), (1,8), (3,8)}$, then does this satisfy the requirement of being onto?
Basically what is piquing my curiosity is if non-function relations can be one-to-one or onto or if being one-to-one and/or onto implies that it must be a function.
abstract-algebra functions
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
$endgroup$
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57
add a comment |
$begingroup$
Let $A = {1,2,3}$ and $B = {7,8,9}$. For the relation between $A$ and $B$ given as a subset of $A times B = {(1,9), (1,7), (3,8)}$, $Atimes B$ is not a function because $1$ appears as the first element in more than one ordered pair. This made me think about the following generalized question.
If a relation is not a function, then it is only a subset of $A$ and $B$ denoted as $Atimes B$. Then can this subset be one-to-one? Can it be onto?
One-to-One?: Even though $Atimes B$ is not a function, I believe it satisfies the requirements of being one-to-one because in $A$, $x_1 = x_2$ only when $x_1 = x_1$.
Onto?: I want to say yes also. The range of $A$ could be the range of $B$. If $A times B = {(1,9), (1,8), (3,8)}$, then does this satisfy the requirement of being onto?
Basically what is piquing my curiosity is if non-function relations can be one-to-one or onto or if being one-to-one and/or onto implies that it must be a function.
abstract-algebra functions
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
$endgroup$
Let $A = {1,2,3}$ and $B = {7,8,9}$. For the relation between $A$ and $B$ given as a subset of $A times B = {(1,9), (1,7), (3,8)}$, $Atimes B$ is not a function because $1$ appears as the first element in more than one ordered pair. This made me think about the following generalized question.
If a relation is not a function, then it is only a subset of $A$ and $B$ denoted as $Atimes B$. Then can this subset be one-to-one? Can it be onto?
One-to-One?: Even though $Atimes B$ is not a function, I believe it satisfies the requirements of being one-to-one because in $A$, $x_1 = x_2$ only when $x_1 = x_1$.
Onto?: I want to say yes also. The range of $A$ could be the range of $B$. If $A times B = {(1,9), (1,8), (3,8)}$, then does this satisfy the requirement of being onto?
Basically what is piquing my curiosity is if non-function relations can be one-to-one or onto or if being one-to-one and/or onto implies that it must be a function.
abstract-algebra functions
abstract-algebra functions
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
asked Mar 17 at 17:50
Evan KimEvan Kim
64919
64919
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57
add a comment |
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer here is yes, relations which are not functions can also be described as injective or surjective. You can read more about it on the wikipedia page here under the section titled "Special types of binary relations," but the definitions are the natural ones:
A relation $Rsubset Atimes B$ is injective if for all $a,a'in A$ and $bin B$, $(a,b)in R$ and $(a',b)in R$ implies $a = a'$.
A relation $Rsubset Atimes B$ is surjective if for each $bin B$, there is some $ain A$ such that $(a,b)in R$.
Some example will hopefully clarify:
If $A = {mathbf{1}}$, a singleton with one element, and $B = {mathbf 1,mathbf 2,mathbf 3}$, then the relation $R = {(mathbf 1,mathbf 1), (mathbf 1,mathbf 2), (mathbf 1,mathbf 3)}$ is surjective. The relation $R$ is not injective, and it is not a function.
If $A = B = Bbb Z$, the set of integers, and $R = emptyset subset Atimes B$ is the empty set, then $R$ is injective (hint: vacuous truth), it is not surjective, and it is a function (vacuous truth again).
If $A = Bbb R$, the set of real numbers, $B = Bbb R_{ge 0}$, the set of non-negative real numbers, and $Rsubset Atimes B$ is the relation $R={(x,x^2): xin Bbb R}$, then $R$ is a surjective function, but it is not injective.
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
$endgroup$
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151830%2fcan-a-relation-that-is-not-a-function-be-one-to-one-or-onto%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer here is yes, relations which are not functions can also be described as injective or surjective. You can read more about it on the wikipedia page here under the section titled "Special types of binary relations," but the definitions are the natural ones:
A relation $Rsubset Atimes B$ is injective if for all $a,a'in A$ and $bin B$, $(a,b)in R$ and $(a',b)in R$ implies $a = a'$.
A relation $Rsubset Atimes B$ is surjective if for each $bin B$, there is some $ain A$ such that $(a,b)in R$.
Some example will hopefully clarify:
If $A = {mathbf{1}}$, a singleton with one element, and $B = {mathbf 1,mathbf 2,mathbf 3}$, then the relation $R = {(mathbf 1,mathbf 1), (mathbf 1,mathbf 2), (mathbf 1,mathbf 3)}$ is surjective. The relation $R$ is not injective, and it is not a function.
If $A = B = Bbb Z$, the set of integers, and $R = emptyset subset Atimes B$ is the empty set, then $R$ is injective (hint: vacuous truth), it is not surjective, and it is a function (vacuous truth again).
If $A = Bbb R$, the set of real numbers, $B = Bbb R_{ge 0}$, the set of non-negative real numbers, and $Rsubset Atimes B$ is the relation $R={(x,x^2): xin Bbb R}$, then $R$ is a surjective function, but it is not injective.
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
$endgroup$
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
add a comment |
$begingroup$
The answer here is yes, relations which are not functions can also be described as injective or surjective. You can read more about it on the wikipedia page here under the section titled "Special types of binary relations," but the definitions are the natural ones:
A relation $Rsubset Atimes B$ is injective if for all $a,a'in A$ and $bin B$, $(a,b)in R$ and $(a',b)in R$ implies $a = a'$.
A relation $Rsubset Atimes B$ is surjective if for each $bin B$, there is some $ain A$ such that $(a,b)in R$.
Some example will hopefully clarify:
If $A = {mathbf{1}}$, a singleton with one element, and $B = {mathbf 1,mathbf 2,mathbf 3}$, then the relation $R = {(mathbf 1,mathbf 1), (mathbf 1,mathbf 2), (mathbf 1,mathbf 3)}$ is surjective. The relation $R$ is not injective, and it is not a function.
If $A = B = Bbb Z$, the set of integers, and $R = emptyset subset Atimes B$ is the empty set, then $R$ is injective (hint: vacuous truth), it is not surjective, and it is a function (vacuous truth again).
If $A = Bbb R$, the set of real numbers, $B = Bbb R_{ge 0}$, the set of non-negative real numbers, and $Rsubset Atimes B$ is the relation $R={(x,x^2): xin Bbb R}$, then $R$ is a surjective function, but it is not injective.
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
$endgroup$
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
add a comment |
$begingroup$
The answer here is yes, relations which are not functions can also be described as injective or surjective. You can read more about it on the wikipedia page here under the section titled "Special types of binary relations," but the definitions are the natural ones:
A relation $Rsubset Atimes B$ is injective if for all $a,a'in A$ and $bin B$, $(a,b)in R$ and $(a',b)in R$ implies $a = a'$.
A relation $Rsubset Atimes B$ is surjective if for each $bin B$, there is some $ain A$ such that $(a,b)in R$.
Some example will hopefully clarify:
If $A = {mathbf{1}}$, a singleton with one element, and $B = {mathbf 1,mathbf 2,mathbf 3}$, then the relation $R = {(mathbf 1,mathbf 1), (mathbf 1,mathbf 2), (mathbf 1,mathbf 3)}$ is surjective. The relation $R$ is not injective, and it is not a function.
If $A = B = Bbb Z$, the set of integers, and $R = emptyset subset Atimes B$ is the empty set, then $R$ is injective (hint: vacuous truth), it is not surjective, and it is a function (vacuous truth again).
If $A = Bbb R$, the set of real numbers, $B = Bbb R_{ge 0}$, the set of non-negative real numbers, and $Rsubset Atimes B$ is the relation $R={(x,x^2): xin Bbb R}$, then $R$ is a surjective function, but it is not injective.
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
$endgroup$
The answer here is yes, relations which are not functions can also be described as injective or surjective. You can read more about it on the wikipedia page here under the section titled "Special types of binary relations," but the definitions are the natural ones:
A relation $Rsubset Atimes B$ is injective if for all $a,a'in A$ and $bin B$, $(a,b)in R$ and $(a',b)in R$ implies $a = a'$.
A relation $Rsubset Atimes B$ is surjective if for each $bin B$, there is some $ain A$ such that $(a,b)in R$.
Some example will hopefully clarify:
If $A = {mathbf{1}}$, a singleton with one element, and $B = {mathbf 1,mathbf 2,mathbf 3}$, then the relation $R = {(mathbf 1,mathbf 1), (mathbf 1,mathbf 2), (mathbf 1,mathbf 3)}$ is surjective. The relation $R$ is not injective, and it is not a function.
If $A = B = Bbb Z$, the set of integers, and $R = emptyset subset Atimes B$ is the empty set, then $R$ is injective (hint: vacuous truth), it is not surjective, and it is a function (vacuous truth again).
If $A = Bbb R$, the set of real numbers, $B = Bbb R_{ge 0}$, the set of non-negative real numbers, and $Rsubset Atimes B$ is the relation $R={(x,x^2): xin Bbb R}$, then $R$ is a surjective function, but it is not injective.
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
edited Mar 17 at 18:13
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
answered Mar 17 at 17:58
Alex OrtizAlex Ortiz
11.3k21441
11.3k21441
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
add a comment |
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
$begingroup$
ahhh great! So in a way, a function is basically a subset of a binary relation!
$endgroup$
– Evan Kim
Mar 17 at 18:02
1
1
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
@EvanKim: Functions are a special case of relations, which are merely subsets of the Cartesian product $Atimes B$. The defining feature of functions is that corresponding to each element of the domain, is a unique element in the codomain.
$endgroup$
– Alex Ortiz
Mar 17 at 18:08
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
$begingroup$
Great, thanks for the edit too. I will be sure to read more about binary relations and review your examples again
$endgroup$
– Evan Kim
Mar 17 at 18:23
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151830%2fcan-a-relation-that-is-not-a-function-be-one-to-one-or-onto%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Your notation is a bit confused. A relation is a subset $R$ of the Cartesian product $Atimes B$. In your example, $|Atimes B| = 3times 3 = 9$, but the relation you are describing is only a subset of that set of $9$ elements.
$endgroup$
– Alex Ortiz
Mar 17 at 17:52
$begingroup$
Hmm. This is how my professor wrote the question down. "For each relation between $A$ and $B$ given as a subset of $Atimes B$, decide whether it is a function mapping $A$ into $B$. I see that $|Atimes B|$ should equal 9, but maybe my professor only wanted a subset of the relation?
$endgroup$
– Evan Kim
Mar 17 at 17:57