Show that a function is injective or surjective when the following condition holds.
$begingroup$
I would like to prove the following two statements.
Let $f:Xto Y$ be a function. If $f^{-1}$($f(A))= $$A$ for every $Asubset$ $X$, then $f$ is one-to-one (or injective).
Let $f:Xto Y$ be a function. If $B= $$f(f^{-1}(B))$ for every $B$ $subset$ $Y$, then $f$ is onto (or surjective).
I proved the converse of each statement on my own, but could not prove the aforementioned parts. I would very much appreciate it if I could get some help.
functions
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
$endgroup$
add a comment |
$begingroup$
I would like to prove the following two statements.
Let $f:Xto Y$ be a function. If $f^{-1}$($f(A))= $$A$ for every $Asubset$ $X$, then $f$ is one-to-one (or injective).
Let $f:Xto Y$ be a function. If $B= $$f(f^{-1}(B))$ for every $B$ $subset$ $Y$, then $f$ is onto (or surjective).
I proved the converse of each statement on my own, but could not prove the aforementioned parts. I would very much appreciate it if I could get some help.
functions
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
$endgroup$
add a comment |
$begingroup$
I would like to prove the following two statements.
Let $f:Xto Y$ be a function. If $f^{-1}$($f(A))= $$A$ for every $Asubset$ $X$, then $f$ is one-to-one (or injective).
Let $f:Xto Y$ be a function. If $B= $$f(f^{-1}(B))$ for every $B$ $subset$ $Y$, then $f$ is onto (or surjective).
I proved the converse of each statement on my own, but could not prove the aforementioned parts. I would very much appreciate it if I could get some help.
functions
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
$endgroup$
I would like to prove the following two statements.
Let $f:Xto Y$ be a function. If $f^{-1}$($f(A))= $$A$ for every $Asubset$ $X$, then $f$ is one-to-one (or injective).
Let $f:Xto Y$ be a function. If $B= $$f(f^{-1}(B))$ for every $B$ $subset$ $Y$, then $f$ is onto (or surjective).
I proved the converse of each statement on my own, but could not prove the aforementioned parts. I would very much appreciate it if I could get some help.
functions
functions
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
asked Apr 7 '17 at 12:33
Tim LeeTim Lee
395
395
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To show that $f$ is injective, let $f(y) =f(x)$.
By definition, given a set $S subset Y$, $f^{-1}(S) = { x in X : f(x) in S}$
Then, if $A = { x }$, we see that $f^{-1} (f(A))= f^{-1}(f(x)) = {x}$, so by definition of $f^{-1}$, since $f(y) = f(x) in f(A), y in f^{-1} (f(x)) = { x}$, so $y=x$. This shows injectivity.
For surjectivity, let $y in Y$. Then, if $B = {y}$, we know that ${ y } = f(f^{-1} (y))$.
Suppose $f^{-1} (y) = emptyset$. Then, $f(f^{-1}(y)) = emptyset$, which is a contradiction. Hence, $f^{-1} (y)$ is non-empty, giving us the conclusion that there is $x$ such that $f(x) =y$. Hence, $f$ is surjective.
The converses of the statements are also easy.
Alternately, by contradiction :
Suppose $f$ is not one-one. Then, there exist $x neq y$ with $f(x) = f(y) =z$. Then, ${x,y} subsetneq f^{-1}(f({x})) neq {x}$, a contradiction.
Suppose $f$ is not onto. Let $b$ be such that there is no $x$ with $f(x) = b$. Of course, $f^{-1}(b) = emptyset$, so $f(f^{-1}({b})) = emptyset neq {b}$, a contradiction.
For the converses :
Suppose $f$ is injective, then we want to show that $f^{-1}(f(A)) = A$ for all $A$. Let $x in A$. Then, $x in f^{-1}(f(x)) subset f^{-1}(f(A))$, so $A subset f^{-1}(f(A))$. Conversely, $y in f^{-1}(f(A))$ implies that $z = f(y) in f(A)$. If $y in A^c$, then $f(y) in f(A^c)$, which is disjoint from $f(A)$ since $A^c$ is disjoint from $A$, and this is not possible. So $y in A$.
For surjectivity too, something similar can be done.
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
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%2f2222307%2fshow-that-a-function-is-injective-or-surjective-when-the-following-condition-hol%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
$begingroup$
To show that $f$ is injective, let $f(y) =f(x)$.
By definition, given a set $S subset Y$, $f^{-1}(S) = { x in X : f(x) in S}$
Then, if $A = { x }$, we see that $f^{-1} (f(A))= f^{-1}(f(x)) = {x}$, so by definition of $f^{-1}$, since $f(y) = f(x) in f(A), y in f^{-1} (f(x)) = { x}$, so $y=x$. This shows injectivity.
For surjectivity, let $y in Y$. Then, if $B = {y}$, we know that ${ y } = f(f^{-1} (y))$.
Suppose $f^{-1} (y) = emptyset$. Then, $f(f^{-1}(y)) = emptyset$, which is a contradiction. Hence, $f^{-1} (y)$ is non-empty, giving us the conclusion that there is $x$ such that $f(x) =y$. Hence, $f$ is surjective.
The converses of the statements are also easy.
Alternately, by contradiction :
Suppose $f$ is not one-one. Then, there exist $x neq y$ with $f(x) = f(y) =z$. Then, ${x,y} subsetneq f^{-1}(f({x})) neq {x}$, a contradiction.
Suppose $f$ is not onto. Let $b$ be such that there is no $x$ with $f(x) = b$. Of course, $f^{-1}(b) = emptyset$, so $f(f^{-1}({b})) = emptyset neq {b}$, a contradiction.
For the converses :
Suppose $f$ is injective, then we want to show that $f^{-1}(f(A)) = A$ for all $A$. Let $x in A$. Then, $x in f^{-1}(f(x)) subset f^{-1}(f(A))$, so $A subset f^{-1}(f(A))$. Conversely, $y in f^{-1}(f(A))$ implies that $z = f(y) in f(A)$. If $y in A^c$, then $f(y) in f(A^c)$, which is disjoint from $f(A)$ since $A^c$ is disjoint from $A$, and this is not possible. So $y in A$.
For surjectivity too, something similar can be done.
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
add a comment |
$begingroup$
To show that $f$ is injective, let $f(y) =f(x)$.
By definition, given a set $S subset Y$, $f^{-1}(S) = { x in X : f(x) in S}$
Then, if $A = { x }$, we see that $f^{-1} (f(A))= f^{-1}(f(x)) = {x}$, so by definition of $f^{-1}$, since $f(y) = f(x) in f(A), y in f^{-1} (f(x)) = { x}$, so $y=x$. This shows injectivity.
For surjectivity, let $y in Y$. Then, if $B = {y}$, we know that ${ y } = f(f^{-1} (y))$.
Suppose $f^{-1} (y) = emptyset$. Then, $f(f^{-1}(y)) = emptyset$, which is a contradiction. Hence, $f^{-1} (y)$ is non-empty, giving us the conclusion that there is $x$ such that $f(x) =y$. Hence, $f$ is surjective.
The converses of the statements are also easy.
Alternately, by contradiction :
Suppose $f$ is not one-one. Then, there exist $x neq y$ with $f(x) = f(y) =z$. Then, ${x,y} subsetneq f^{-1}(f({x})) neq {x}$, a contradiction.
Suppose $f$ is not onto. Let $b$ be such that there is no $x$ with $f(x) = b$. Of course, $f^{-1}(b) = emptyset$, so $f(f^{-1}({b})) = emptyset neq {b}$, a contradiction.
For the converses :
Suppose $f$ is injective, then we want to show that $f^{-1}(f(A)) = A$ for all $A$. Let $x in A$. Then, $x in f^{-1}(f(x)) subset f^{-1}(f(A))$, so $A subset f^{-1}(f(A))$. Conversely, $y in f^{-1}(f(A))$ implies that $z = f(y) in f(A)$. If $y in A^c$, then $f(y) in f(A^c)$, which is disjoint from $f(A)$ since $A^c$ is disjoint from $A$, and this is not possible. So $y in A$.
For surjectivity too, something similar can be done.
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
add a comment |
$begingroup$
To show that $f$ is injective, let $f(y) =f(x)$.
By definition, given a set $S subset Y$, $f^{-1}(S) = { x in X : f(x) in S}$
Then, if $A = { x }$, we see that $f^{-1} (f(A))= f^{-1}(f(x)) = {x}$, so by definition of $f^{-1}$, since $f(y) = f(x) in f(A), y in f^{-1} (f(x)) = { x}$, so $y=x$. This shows injectivity.
For surjectivity, let $y in Y$. Then, if $B = {y}$, we know that ${ y } = f(f^{-1} (y))$.
Suppose $f^{-1} (y) = emptyset$. Then, $f(f^{-1}(y)) = emptyset$, which is a contradiction. Hence, $f^{-1} (y)$ is non-empty, giving us the conclusion that there is $x$ such that $f(x) =y$. Hence, $f$ is surjective.
The converses of the statements are also easy.
Alternately, by contradiction :
Suppose $f$ is not one-one. Then, there exist $x neq y$ with $f(x) = f(y) =z$. Then, ${x,y} subsetneq f^{-1}(f({x})) neq {x}$, a contradiction.
Suppose $f$ is not onto. Let $b$ be such that there is no $x$ with $f(x) = b$. Of course, $f^{-1}(b) = emptyset$, so $f(f^{-1}({b})) = emptyset neq {b}$, a contradiction.
For the converses :
Suppose $f$ is injective, then we want to show that $f^{-1}(f(A)) = A$ for all $A$. Let $x in A$. Then, $x in f^{-1}(f(x)) subset f^{-1}(f(A))$, so $A subset f^{-1}(f(A))$. Conversely, $y in f^{-1}(f(A))$ implies that $z = f(y) in f(A)$. If $y in A^c$, then $f(y) in f(A^c)$, which is disjoint from $f(A)$ since $A^c$ is disjoint from $A$, and this is not possible. So $y in A$.
For surjectivity too, something similar can be done.
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
$endgroup$
To show that $f$ is injective, let $f(y) =f(x)$.
By definition, given a set $S subset Y$, $f^{-1}(S) = { x in X : f(x) in S}$
Then, if $A = { x }$, we see that $f^{-1} (f(A))= f^{-1}(f(x)) = {x}$, so by definition of $f^{-1}$, since $f(y) = f(x) in f(A), y in f^{-1} (f(x)) = { x}$, so $y=x$. This shows injectivity.
For surjectivity, let $y in Y$. Then, if $B = {y}$, we know that ${ y } = f(f^{-1} (y))$.
Suppose $f^{-1} (y) = emptyset$. Then, $f(f^{-1}(y)) = emptyset$, which is a contradiction. Hence, $f^{-1} (y)$ is non-empty, giving us the conclusion that there is $x$ such that $f(x) =y$. Hence, $f$ is surjective.
The converses of the statements are also easy.
Alternately, by contradiction :
Suppose $f$ is not one-one. Then, there exist $x neq y$ with $f(x) = f(y) =z$. Then, ${x,y} subsetneq f^{-1}(f({x})) neq {x}$, a contradiction.
Suppose $f$ is not onto. Let $b$ be such that there is no $x$ with $f(x) = b$. Of course, $f^{-1}(b) = emptyset$, so $f(f^{-1}({b})) = emptyset neq {b}$, a contradiction.
For the converses :
Suppose $f$ is injective, then we want to show that $f^{-1}(f(A)) = A$ for all $A$. Let $x in A$. Then, $x in f^{-1}(f(x)) subset f^{-1}(f(A))$, so $A subset f^{-1}(f(A))$. Conversely, $y in f^{-1}(f(A))$ implies that $z = f(y) in f(A)$. If $y in A^c$, then $f(y) in f(A^c)$, which is disjoint from $f(A)$ since $A^c$ is disjoint from $A$, and this is not possible. So $y in A$.
For surjectivity too, something similar can be done.
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
edited Jan 3 at 7:17
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
answered Apr 7 '17 at 12:42
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.1k33577
40.1k33577
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
add a comment |
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
Thanks you so much for your help. However, I have some parts I do not understand and I would appreciate it if you could spare some time for me. In the question, A is an arbitrary subset of X, but can I assume that A is a singleton set consisting of x? (the same question applies to an "arbitrary" subset B of Y which you assumed to be a singleton set consisting of y) Should I think of other possibilities such as a possibility that A (B respectively) is not a singleton set which includes other elements of X (Y respectively)? Thank you for your help in advance.
$endgroup$
– Tim Lee
Apr 7 '17 at 15:10
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
@TimLee Certainly, if $A$ is an arbitrary subset of $X$, then we can choose the subset consisting of just one element. The same applies for $B$ too. You could try to think about what happens if you include more than one element,that would be interesting. However, then I think the statement won't be true, in fact. The point is, if an element $y$ belongs to a singleton ${x}$, then $y=x$ must happen, whereas if you allow $A$ or $B$ to contain more than one element, this allows $y$ to have one of two values, which you cannot control. This is, I think, fatal to the argument.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:06
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
In continuation (and conclusion), the fact you can choose $A,B$ to be singletons is key to the argument for the reason I said above. So, if you fix $A,B$ to have at least two elements, then I think you will find a counterexample to both the above propositions.
$endgroup$
– астон вілла олоф мэллбэрг
Apr 8 '17 at 3:08
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
Thank you so much for your kind reply. Now I understand that letting A and B being singleton sets is key in this question. Thank you again so much for your help.
$endgroup$
– Tim Lee
Apr 8 '17 at 12:50
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
$begingroup$
@TimLee You are welcome!
$endgroup$
– астон вілла олоф мэллбэрг
Apr 9 '17 at 3:36
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%2f2222307%2fshow-that-a-function-is-injective-or-surjective-when-the-following-condition-hol%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
