If $sup A = 5$ and $B = left{ 3a mid a in A right}$ then $sup B = 15$
$begingroup$
Prove that if $A subset mathbb{R}$, $sup A = 5$, and $B = left{ 3a mid a in A right}$, then $sup = 15$.
I tried to do contradiction by assuming the hypothesis and that there is a number $< 15$ that is the supremum of $B$. Then $3a < 15$ is the $sup B$, and then we divide both sides by $3$ to get $a < 5$. To use contradiction for this particular case, I would have to show that if $3a$ is an upper bound for $B$, then that implies that $a$ is an upper bound for $A$ (so that I can say that $a < 5$ is an upper bound for $A$, which is a contradiction to the hypothesis that $5$ is the least upper bound for $A$). But I don't know how to show that $3a$ being an upper bound for $B$ implies that $a$ is an upper bound for $A$. Or am I going about this proof incorrectly?
supremum-and-infimum
edited Dec 15 '14 at 14:39
Did
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
$endgroup$
add a comment |
$begingroup$
Prove that if $A subset mathbb{R}$, $sup A = 5$, and $B = left{ 3a mid a in A right}$, then $sup = 15$.
I tried to do contradiction by assuming the hypothesis and that there is a number $< 15$ that is the supremum of $B$. Then $3a < 15$ is the $sup B$, and then we divide both sides by $3$ to get $a < 5$. To use contradiction for this particular case, I would have to show that if $3a$ is an upper bound for $B$, then that implies that $a$ is an upper bound for $A$ (so that I can say that $a < 5$ is an upper bound for $A$, which is a contradiction to the hypothesis that $5$ is the least upper bound for $A$). But I don't know how to show that $3a$ being an upper bound for $B$ implies that $a$ is an upper bound for $A$. Or am I going about this proof incorrectly?
supremum-and-infimum
edited Dec 15 '14 at 14:39
Did
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
$endgroup$
1
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
1
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38
add a comment |
$begingroup$
Prove that if $A subset mathbb{R}$, $sup A = 5$, and $B = left{ 3a mid a in A right}$, then $sup = 15$.
I tried to do contradiction by assuming the hypothesis and that there is a number $< 15$ that is the supremum of $B$. Then $3a < 15$ is the $sup B$, and then we divide both sides by $3$ to get $a < 5$. To use contradiction for this particular case, I would have to show that if $3a$ is an upper bound for $B$, then that implies that $a$ is an upper bound for $A$ (so that I can say that $a < 5$ is an upper bound for $A$, which is a contradiction to the hypothesis that $5$ is the least upper bound for $A$). But I don't know how to show that $3a$ being an upper bound for $B$ implies that $a$ is an upper bound for $A$. Or am I going about this proof incorrectly?
supremum-and-infimum
edited Dec 15 '14 at 14:39
Did
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
$endgroup$
Prove that if $A subset mathbb{R}$, $sup A = 5$, and $B = left{ 3a mid a in A right}$, then $sup = 15$.
I tried to do contradiction by assuming the hypothesis and that there is a number $< 15$ that is the supremum of $B$. Then $3a < 15$ is the $sup B$, and then we divide both sides by $3$ to get $a < 5$. To use contradiction for this particular case, I would have to show that if $3a$ is an upper bound for $B$, then that implies that $a$ is an upper bound for $A$ (so that I can say that $a < 5$ is an upper bound for $A$, which is a contradiction to the hypothesis that $5$ is the least upper bound for $A$). But I don't know how to show that $3a$ being an upper bound for $B$ implies that $a$ is an upper bound for $A$. Or am I going about this proof incorrectly?
supremum-and-infimum
supremum-and-infimum
edited Dec 15 '14 at 14:39
Did
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
edited Dec 15 '14 at 14:39
Did
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
edited Dec 15 '14 at 14:39
Did
247k23223460
edited Dec 15 '14 at 14:39
Did
247k23223460
edited Dec 15 '14 at 14:39
Did
247k23223460
247k23223460
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
asked Dec 15 '14 at 14:17
mr eyeglassesmr eyeglasses
2,47332048
2,47332048
1
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
1
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38
add a comment |
1
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
1
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38
1
1
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
1
1
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $sup A=5$ then $a_nto 5$ for some $a_nin A$. But since $3a_nin B$ and $3a_nto 3cdot 5=15$, therefore $sup B ge 15$. For $sup B le 15$ notice that any element of $B$ has the form $3a,ain A$ and $3ale 3cdot sup A=15$.
answered Dec 15 '14 at 14:24
MherMher
3,8301432
$endgroup$
add a comment |
$begingroup$
We have
$$forall a in A,; ale sup Aimplies forall a in A,; 3ale 3sup A $$
and
$$forall epsilon>0,; exists ain A ;|; sup A-fracepsilon 3< aimplies 3sup A-epsilon<3a$$
so we have
$$3sup A=sup{3a;|; ain A}$$
$endgroup$
add a comment |
$begingroup$
While $C=${$3$}, it defines $B=Ccdot A={{x cdot y mid x in C, y in A}}$
remembering the theorem: $$sup(A cdot B)= sup(A) cdot sup(B)$$
so:$hspace{2cm}$ $sup(B)=sup(C cdot A)=sup(C) cdot sup(A)=3 cdot 5=15$
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
$endgroup$
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%2f1069147%2fif-sup-a-5-and-b-left-3a-mid-a-in-a-right-then-sup-b-15%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $sup A=5$ then $a_nto 5$ for some $a_nin A$. But since $3a_nin B$ and $3a_nto 3cdot 5=15$, therefore $sup B ge 15$. For $sup B le 15$ notice that any element of $B$ has the form $3a,ain A$ and $3ale 3cdot sup A=15$.
answered Dec 15 '14 at 14:24
MherMher
3,8301432
$endgroup$
add a comment |
$begingroup$
Since $sup A=5$ then $a_nto 5$ for some $a_nin A$. But since $3a_nin B$ and $3a_nto 3cdot 5=15$, therefore $sup B ge 15$. For $sup B le 15$ notice that any element of $B$ has the form $3a,ain A$ and $3ale 3cdot sup A=15$.
answered Dec 15 '14 at 14:24
MherMher
3,8301432
$endgroup$
add a comment |
$begingroup$
Since $sup A=5$ then $a_nto 5$ for some $a_nin A$. But since $3a_nin B$ and $3a_nto 3cdot 5=15$, therefore $sup B ge 15$. For $sup B le 15$ notice that any element of $B$ has the form $3a,ain A$ and $3ale 3cdot sup A=15$.
answered Dec 15 '14 at 14:24
MherMher
3,8301432
$endgroup$
Since $sup A=5$ then $a_nto 5$ for some $a_nin A$. But since $3a_nin B$ and $3a_nto 3cdot 5=15$, therefore $sup B ge 15$. For $sup B le 15$ notice that any element of $B$ has the form $3a,ain A$ and $3ale 3cdot sup A=15$.
answered Dec 15 '14 at 14:24
MherMher
3,8301432
answered Dec 15 '14 at 14:24
MherMher
3,8301432
answered Dec 15 '14 at 14:24
MherMher
3,8301432
answered Dec 15 '14 at 14:24
MherMher
3,8301432
3,8301432
add a comment |
add a comment |
$begingroup$
We have
$$forall a in A,; ale sup Aimplies forall a in A,; 3ale 3sup A $$
and
$$forall epsilon>0,; exists ain A ;|; sup A-fracepsilon 3< aimplies 3sup A-epsilon<3a$$
so we have
$$3sup A=sup{3a;|; ain A}$$
$endgroup$
add a comment |
$begingroup$
We have
$$forall a in A,; ale sup Aimplies forall a in A,; 3ale 3sup A $$
and
$$forall epsilon>0,; exists ain A ;|; sup A-fracepsilon 3< aimplies 3sup A-epsilon<3a$$
so we have
$$3sup A=sup{3a;|; ain A}$$
$endgroup$
add a comment |
$begingroup$
We have
$$forall a in A,; ale sup Aimplies forall a in A,; 3ale 3sup A $$
and
$$forall epsilon>0,; exists ain A ;|; sup A-fracepsilon 3< aimplies 3sup A-epsilon<3a$$
so we have
$$3sup A=sup{3a;|; ain A}$$
$endgroup$
We have
$$forall a in A,; ale sup Aimplies forall a in A,; 3ale 3sup A $$
and
$$forall epsilon>0,; exists ain A ;|; sup A-fracepsilon 3< aimplies 3sup A-epsilon<3a$$
so we have
$$3sup A=sup{3a;|; ain A}$$
answered Dec 15 '14 at 14:24
user63181
add a comment |
add a comment |
$begingroup$
While $C=${$3$}, it defines $B=Ccdot A={{x cdot y mid x in C, y in A}}$
remembering the theorem: $$sup(A cdot B)= sup(A) cdot sup(B)$$
so:$hspace{2cm}$ $sup(B)=sup(C cdot A)=sup(C) cdot sup(A)=3 cdot 5=15$
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
$endgroup$
add a comment |
$begingroup$
While $C=${$3$}, it defines $B=Ccdot A={{x cdot y mid x in C, y in A}}$
remembering the theorem: $$sup(A cdot B)= sup(A) cdot sup(B)$$
so:$hspace{2cm}$ $sup(B)=sup(C cdot A)=sup(C) cdot sup(A)=3 cdot 5=15$
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
$endgroup$
add a comment |
$begingroup$
While $C=${$3$}, it defines $B=Ccdot A={{x cdot y mid x in C, y in A}}$
remembering the theorem: $$sup(A cdot B)= sup(A) cdot sup(B)$$
so:$hspace{2cm}$ $sup(B)=sup(C cdot A)=sup(C) cdot sup(A)=3 cdot 5=15$
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
$endgroup$
While $C=${$3$}, it defines $B=Ccdot A={{x cdot y mid x in C, y in A}}$
remembering the theorem: $$sup(A cdot B)= sup(A) cdot sup(B)$$
so:$hspace{2cm}$ $sup(B)=sup(C cdot A)=sup(C) cdot sup(A)=3 cdot 5=15$
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
answered Jan 15 at 4:10
Francisco SalazarFrancisco Salazar
85
85
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%2f1069147%2fif-sup-a-5-and-b-left-3a-mid-a-in-a-right-then-sup-b-15%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
$begingroup$
Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false.
$endgroup$
– John Hughes
Dec 15 '14 at 14:19
1
$begingroup$
Related.
$endgroup$
– Git Gud
Dec 15 '14 at 14:28
$begingroup$
Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me
$endgroup$
– mr eyeglasses
Dec 15 '14 at 14:38