How to Prove The Complement Of The Domain Is Complement Of The Image If f Is Bijective
$begingroup$
It seems true that $f(overline{X}) = overline{f(X)}$ for $f:Arightarrow B$ and $X$ is any subset of $A$ if and only if $f$ is bijective.But I couldn't write it as a formal way like epsilon argument.It makes sense to me but the trouble I have is with the formal prove.
functions discrete-mathematics theorem-provers
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
$endgroup$
add a comment |
$begingroup$
It seems true that $f(overline{X}) = overline{f(X)}$ for $f:Arightarrow B$ and $X$ is any subset of $A$ if and only if $f$ is bijective.But I couldn't write it as a formal way like epsilon argument.It makes sense to me but the trouble I have is with the formal prove.
functions discrete-mathematics theorem-provers
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
$endgroup$
1
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33
add a comment |
$begingroup$
It seems true that $f(overline{X}) = overline{f(X)}$ for $f:Arightarrow B$ and $X$ is any subset of $A$ if and only if $f$ is bijective.But I couldn't write it as a formal way like epsilon argument.It makes sense to me but the trouble I have is with the formal prove.
functions discrete-mathematics theorem-provers
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
$endgroup$
It seems true that $f(overline{X}) = overline{f(X)}$ for $f:Arightarrow B$ and $X$ is any subset of $A$ if and only if $f$ is bijective.But I couldn't write it as a formal way like epsilon argument.It makes sense to me but the trouble I have is with the formal prove.
functions discrete-mathematics theorem-provers
functions discrete-mathematics theorem-provers
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
edited Dec 9 '18 at 13:25
selman özlyn
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
asked Dec 9 '18 at 12:53
selman özlynselman özlyn
12
12
1
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33
add a comment |
1
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33
1
1
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A rightarrow B$. We have the following elementary propositions:
$f$ is injective if and only if
$$(forall X)(X subseteq A implies f(complement_{A}X) subseteq complement_{B}f(X))$$
Proof : Assuming first the injectivity of $f$, consider arbitrary $X subseteq A$ and $y in f(complement_{A}X)$, such that $y=f(x)$ with $x in A setminus X$. If we were to assume by contradiction that $y in f(X)$ it would entail that $y=f(t)$ for a certain $t in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t in X$ in contradiction to $x notin X$. Hence $y in B setminus f(X)$ and the inclusion is established.
Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y in A$ with $x neq y$. This means that $y in A setminus {x}$ and thus by our assumption
$$f({y})={f(y)} subseteq f(A setminus {x})subseteq B setminus f({x})=B setminus {f(x)}$$
from which we infer that $f(x) neq f(y)$ and conclude $f$ is indeed injective. $Box$
$f$ is bijective if and only if:
$$(forall X)(X subseteq A implies f(complement_{A}X)=complement_{B}f(X))$$
Proof : Assume first $f$ is bijective and consider arbitrary $X subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:
$$(forall X)(X subseteq A implies f(A) setminus f(X) subseteq f(A setminus X))$$
and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.
To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= emptyset$, which yields:
$$f(A setminus emptyset)=f(A)=B setminus f(emptyset)=B setminus emptyset =B$$
This concludes the proof. $Box$
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
Your Answer
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%2f3032357%2fhow-to-prove-the-complement-of-the-domain-is-complement-of-the-image-if-f-is-bij%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
$begingroup$
Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A rightarrow B$. We have the following elementary propositions:
$f$ is injective if and only if
$$(forall X)(X subseteq A implies f(complement_{A}X) subseteq complement_{B}f(X))$$
Proof : Assuming first the injectivity of $f$, consider arbitrary $X subseteq A$ and $y in f(complement_{A}X)$, such that $y=f(x)$ with $x in A setminus X$. If we were to assume by contradiction that $y in f(X)$ it would entail that $y=f(t)$ for a certain $t in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t in X$ in contradiction to $x notin X$. Hence $y in B setminus f(X)$ and the inclusion is established.
Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y in A$ with $x neq y$. This means that $y in A setminus {x}$ and thus by our assumption
$$f({y})={f(y)} subseteq f(A setminus {x})subseteq B setminus f({x})=B setminus {f(x)}$$
from which we infer that $f(x) neq f(y)$ and conclude $f$ is indeed injective. $Box$
$f$ is bijective if and only if:
$$(forall X)(X subseteq A implies f(complement_{A}X)=complement_{B}f(X))$$
Proof : Assume first $f$ is bijective and consider arbitrary $X subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:
$$(forall X)(X subseteq A implies f(A) setminus f(X) subseteq f(A setminus X))$$
and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.
To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= emptyset$, which yields:
$$f(A setminus emptyset)=f(A)=B setminus f(emptyset)=B setminus emptyset =B$$
This concludes the proof. $Box$
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A rightarrow B$. We have the following elementary propositions:
$f$ is injective if and only if
$$(forall X)(X subseteq A implies f(complement_{A}X) subseteq complement_{B}f(X))$$
Proof : Assuming first the injectivity of $f$, consider arbitrary $X subseteq A$ and $y in f(complement_{A}X)$, such that $y=f(x)$ with $x in A setminus X$. If we were to assume by contradiction that $y in f(X)$ it would entail that $y=f(t)$ for a certain $t in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t in X$ in contradiction to $x notin X$. Hence $y in B setminus f(X)$ and the inclusion is established.
Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y in A$ with $x neq y$. This means that $y in A setminus {x}$ and thus by our assumption
$$f({y})={f(y)} subseteq f(A setminus {x})subseteq B setminus f({x})=B setminus {f(x)}$$
from which we infer that $f(x) neq f(y)$ and conclude $f$ is indeed injective. $Box$
$f$ is bijective if and only if:
$$(forall X)(X subseteq A implies f(complement_{A}X)=complement_{B}f(X))$$
Proof : Assume first $f$ is bijective and consider arbitrary $X subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:
$$(forall X)(X subseteq A implies f(A) setminus f(X) subseteq f(A setminus X))$$
and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.
To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= emptyset$, which yields:
$$f(A setminus emptyset)=f(A)=B setminus f(emptyset)=B setminus emptyset =B$$
This concludes the proof. $Box$
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
$endgroup$
add a comment |
$begingroup$
Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A rightarrow B$. We have the following elementary propositions:
$f$ is injective if and only if
$$(forall X)(X subseteq A implies f(complement_{A}X) subseteq complement_{B}f(X))$$
Proof : Assuming first the injectivity of $f$, consider arbitrary $X subseteq A$ and $y in f(complement_{A}X)$, such that $y=f(x)$ with $x in A setminus X$. If we were to assume by contradiction that $y in f(X)$ it would entail that $y=f(t)$ for a certain $t in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t in X$ in contradiction to $x notin X$. Hence $y in B setminus f(X)$ and the inclusion is established.
Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y in A$ with $x neq y$. This means that $y in A setminus {x}$ and thus by our assumption
$$f({y})={f(y)} subseteq f(A setminus {x})subseteq B setminus f({x})=B setminus {f(x)}$$
from which we infer that $f(x) neq f(y)$ and conclude $f$ is indeed injective. $Box$
$f$ is bijective if and only if:
$$(forall X)(X subseteq A implies f(complement_{A}X)=complement_{B}f(X))$$
Proof : Assume first $f$ is bijective and consider arbitrary $X subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:
$$(forall X)(X subseteq A implies f(A) setminus f(X) subseteq f(A setminus X))$$
and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.
To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= emptyset$, which yields:
$$f(A setminus emptyset)=f(A)=B setminus f(emptyset)=B setminus emptyset =B$$
This concludes the proof. $Box$
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
$endgroup$
Let us formulate matters more precisely: consider two arbitrary sets $A, B$ and a map $f: A rightarrow B$. We have the following elementary propositions:
$f$ is injective if and only if
$$(forall X)(X subseteq A implies f(complement_{A}X) subseteq complement_{B}f(X))$$
Proof : Assuming first the injectivity of $f$, consider arbitrary $X subseteq A$ and $y in f(complement_{A}X)$, such that $y=f(x)$ with $x in A setminus X$. If we were to assume by contradiction that $y in f(X)$ it would entail that $y=f(t)$ for a certain $t in X$; as $y=f(x)=f(t)$ and $f$ is injective, we could conclude $x=t in X$ in contradiction to $x notin X$. Hence $y in B setminus f(X)$ and the inclusion is established.
Assuming conversely that the stated inclusion holds for any subset $X$, let us consider arbitrary $x, y in A$ with $x neq y$. This means that $y in A setminus {x}$ and thus by our assumption
$$f({y})={f(y)} subseteq f(A setminus {x})subseteq B setminus f({x})=B setminus {f(x)}$$
from which we infer that $f(x) neq f(y)$ and conclude $f$ is indeed injective. $Box$
$f$ is bijective if and only if:
$$(forall X)(X subseteq A implies f(complement_{A}X)=complement_{B}f(X))$$
Proof : Assume first $f$ is bijective and consider arbitrary $X subseteq A$. Bijectivity comprises injectivity and thus the previous result entails that we have a valid inclusion in the direction specified above. As for the reverse inclusion we quote the following general result, valid for any function without any special hypotheses:
$$(forall X)(X subseteq A implies f(A) setminus f(X) subseteq f(A setminus X))$$
and we bear in mind that since $f$ is also assumed to be surjective, we automatically have $f(A)=B$.
To establish the reverse implication, assuming the given relation of equality for any subset $X$ we first infer by 1. that $f$ is injective; its surjectivity we derive by considering the particular case $X= emptyset$, which yields:
$$f(A setminus emptyset)=f(A)=B setminus f(emptyset)=B setminus emptyset =B$$
This concludes the proof. $Box$
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
answered Dec 9 '18 at 13:33
ΑΘΩΑΘΩ
2463
2463
add a comment |
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%2f3032357%2fhow-to-prove-the-complement-of-the-domain-is-complement-of-the-image-if-f-is-bij%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
1
$begingroup$
Of course not ! Take $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}$. Is surjective : $[0,1]to mathbb Qcap [0,1]$ but $fleft(overline{[0,1]cap mathbb Q}right)neq overline{f([0,1]cap mathbb Q)}$
$endgroup$
– Surb
Dec 9 '18 at 13:02
$begingroup$
What if it is bijective it is surely true. Let me change the question then
$endgroup$
– selman özlyn
Dec 9 '18 at 13:12
$begingroup$
Now it's not : $f(x)=xboldsymbol 1_{mathbb Qcap [0,1]}-xboldsymbol 1_{mathbb Rsetminus mathbb Qcap [0,1]}$. It's bijective $[0,1]to (mathbb Qcap [0,1])cup((mathbb Rsetminus mathbb Q)cap (0,-1])$ but $f(overline{[0,1]cap mathbb Q})neq overline{f([0,1]cap mathbb Q)}$. What is exactely your exercise ?
$endgroup$
– Surb
Dec 9 '18 at 13:33