Mixing
Let $A$ be bounded below, and $B = {b in R : b$ is a lower bound for $A}$. Show that $sup B = inf A$.
$begingroup$
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = {b in R : b$ is a lower bound for $A}$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $inf{A}$. Then $B={b in mathbb R:b$ is a lower bound for $A}$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup {B}=inf {A}$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
$endgroup$
|
show 6 more comments
$begingroup$
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = {b in R : b$ is a lower bound for $A}$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $inf{A}$. Then $B={b in mathbb R:b$ is a lower bound for $A}$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup {B}=inf {A}$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
$endgroup$
1
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
1
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
1
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
1
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
2
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23
|
show 6 more comments
$begingroup$
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = {b in R : b$ is a lower bound for $A}$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $inf{A}$. Then $B={b in mathbb R:b$ is a lower bound for $A}$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup {B}=inf {A}$.
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
$endgroup$
This is Exercise 1.3.3 (a) from "Understanding Analysis" by Stephen Abbot, , page 17:
Let $A$ be bounded below, and define $B = {b in R : b$ is a lower bound for $A}$. Show that $sup B = inf A$.
Please check my proof:
Suppose $A$ be bound below, it exist number $Nleq n$ for every $nin mathbb R$ in the set and $N$ is $inf{A}$. Then $B={b in mathbb R:b$ is a lower bound for $A}$. It exist number $b$ in the set, since it contains only lower bound of $A$ then it has number $bleq M$ for every $b$ and $M$ is $sup$ of $B$. But $b$ is lower bound of $A$ and $A$ has $N$ as $sup$, $N$ is in $B$. $N$ is automatically is sup of $b$ therefore $M=N$ or $sup {B}=inf {A}$.
real-analysis proof-verification supremum-and-infimum
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
edited Jan 16 '17 at 12:20
Martin Sleziak
44.9k10121274
44.9k10121274
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
asked Jan 16 '17 at 10:50
Lingnoi401Lingnoi401
925520
925520
1
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
1
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
1
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
1
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
2
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23
|
show 6 more comments
1
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
1
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
1
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
1
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
2
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23
1
1
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
1
1
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
1
1
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
1
1
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
2
2
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 '18 at 11:39
BentapairBentapair
1
$endgroup$
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%2f2099966%2flet-a-be-bounded-below-and-b-b-in-r-b-is-a-lower-bound-for-a-sh%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
$begingroup$
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 '18 at 11:39
BentapairBentapair
1
$endgroup$
add a comment |
$begingroup$
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 '18 at 11:39
BentapairBentapair
1
$endgroup$
add a comment |
$begingroup$
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 '18 at 11:39
BentapairBentapair
1
$endgroup$
$inf A$ is defined as the biggest lower bound of $A$. So it is in particular a lower bound of $A$. But that means it is an element of $B$, which is defined as the set of all lower bounds of $A$. Thus $inf Ain B$. But it is also the biggest lower bound of $A$, so for every $bin B$ follows $bleqinf A$. This gives us $inf A=max B$. But if a set has a maximum, this maximum is also its supremum. Thus $inf A=max B=sup B$.
answered Aug 14 '18 at 11:39
BentapairBentapair
1
answered Aug 14 '18 at 11:39
BentapairBentapair
1
answered Aug 14 '18 at 11:39
BentapairBentapair
1
answered Aug 14 '18 at 11:39
BentapairBentapair
1
1
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%2f2099966%2flet-a-be-bounded-below-and-b-b-in-r-b-is-a-lower-bound-for-a-sh%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
$begingroup$
wait ,why edited my question ?
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:00
1
$begingroup$
@juniven exercise from Understanding Analysis by Stephen Abbot
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:08
1
$begingroup$
it is in page 17
$endgroup$
– Lingnoi401
Jan 16 '17 at 11:09
1
$begingroup$
There I see Exercise 1.3.3 (a) Thank you.
$endgroup$
– ΘΣΦGenSan
Jan 16 '17 at 11:13
2
$begingroup$
You question was edited as your typesetting was a bit unusual. Moreover, the body of the post should be self-contained and not depend on the content of the title (which in turn was/is a tad on the too complicated side for a title).
$endgroup$
– quid♦
Jan 16 '17 at 12:23