Show that if $inf(A^+)=a>0$ then $ain A$ and $A={za;zin mathbb{Z}}$
$begingroup$
$A$ is a set such that $x,yin ARightarrow$ $x-yin A$, and $A^+$ is a subset of $A$ which contains only its positive elements.
I was able to successfully show that if $nain A$, $ninmathbb{Z}, n neq 0$ then $(na,na pm a)cap A=emptyset$ and that if $infA^+=0$ then $A$ is dense in $mathbb{R}$.
Not sure if the first the latter is useful in this context but I am leaving it here in case it is (maybe consider set $A-a$?).
Now, the only thing left for me to prove is that all $zain A$ for $zin mathbb{Z}$. From what I've observed, it suffices to show that $ain A$, but I have been stuck trying to prove that for a while with no success whatsoever.
If anyone could offer some help it would be highly appreciated. Please do not just give out the answer though. Some tip/insight to get me on the right track would be more interesting.
real-analysis
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
$endgroup$
add a comment |
$begingroup$
$A$ is a set such that $x,yin ARightarrow$ $x-yin A$, and $A^+$ is a subset of $A$ which contains only its positive elements.
I was able to successfully show that if $nain A$, $ninmathbb{Z}, n neq 0$ then $(na,na pm a)cap A=emptyset$ and that if $infA^+=0$ then $A$ is dense in $mathbb{R}$.
Not sure if the first the latter is useful in this context but I am leaving it here in case it is (maybe consider set $A-a$?).
Now, the only thing left for me to prove is that all $zain A$ for $zin mathbb{Z}$. From what I've observed, it suffices to show that $ain A$, but I have been stuck trying to prove that for a while with no success whatsoever.
If anyone could offer some help it would be highly appreciated. Please do not just give out the answer though. Some tip/insight to get me on the right track would be more interesting.
real-analysis
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
$endgroup$
add a comment |
$begingroup$
$A$ is a set such that $x,yin ARightarrow$ $x-yin A$, and $A^+$ is a subset of $A$ which contains only its positive elements.
I was able to successfully show that if $nain A$, $ninmathbb{Z}, n neq 0$ then $(na,na pm a)cap A=emptyset$ and that if $infA^+=0$ then $A$ is dense in $mathbb{R}$.
Not sure if the first the latter is useful in this context but I am leaving it here in case it is (maybe consider set $A-a$?).
Now, the only thing left for me to prove is that all $zain A$ for $zin mathbb{Z}$. From what I've observed, it suffices to show that $ain A$, but I have been stuck trying to prove that for a while with no success whatsoever.
If anyone could offer some help it would be highly appreciated. Please do not just give out the answer though. Some tip/insight to get me on the right track would be more interesting.
real-analysis
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
$endgroup$
$A$ is a set such that $x,yin ARightarrow$ $x-yin A$, and $A^+$ is a subset of $A$ which contains only its positive elements.
I was able to successfully show that if $nain A$, $ninmathbb{Z}, n neq 0$ then $(na,na pm a)cap A=emptyset$ and that if $infA^+=0$ then $A$ is dense in $mathbb{R}$.
Not sure if the first the latter is useful in this context but I am leaving it here in case it is (maybe consider set $A-a$?).
Now, the only thing left for me to prove is that all $zain A$ for $zin mathbb{Z}$. From what I've observed, it suffices to show that $ain A$, but I have been stuck trying to prove that for a while with no success whatsoever.
If anyone could offer some help it would be highly appreciated. Please do not just give out the answer though. Some tip/insight to get me on the right track would be more interesting.
real-analysis
real-analysis
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
asked Dec 10 '18 at 15:11
PolygonsPolygons
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe: $A$ is a group with addition: Assume it is nonempty so $exists xin A$.
i) $x-x=0 in A$
ii) If $xin A$ then $0-x in A$
iii) If $x,yin A$ then $x+y=x-(-y)in A$
Now the statement is basically saying discrete subgroups of $mathbb{R}$ is cyclic:
Suppose your chosen $anotin A$. Then there must be a sequence in $A$: $a_ndownarrow a$ by definition of $inf$. Now consider the sequence $(a_n-a_m)_{m,n}$ to get a contradiction. (A convergent sequence is Cauchy)
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
$endgroup$
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%2f3034037%2fshow-that-if-infa-a0-then-a-in-a-and-a-zaz-in-mathbbz%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
$begingroup$
Observe: $A$ is a group with addition: Assume it is nonempty so $exists xin A$.
i) $x-x=0 in A$
ii) If $xin A$ then $0-x in A$
iii) If $x,yin A$ then $x+y=x-(-y)in A$
Now the statement is basically saying discrete subgroups of $mathbb{R}$ is cyclic:
Suppose your chosen $anotin A$. Then there must be a sequence in $A$: $a_ndownarrow a$ by definition of $inf$. Now consider the sequence $(a_n-a_m)_{m,n}$ to get a contradiction. (A convergent sequence is Cauchy)
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
$endgroup$
add a comment |
$begingroup$
Observe: $A$ is a group with addition: Assume it is nonempty so $exists xin A$.
i) $x-x=0 in A$
ii) If $xin A$ then $0-x in A$
iii) If $x,yin A$ then $x+y=x-(-y)in A$
Now the statement is basically saying discrete subgroups of $mathbb{R}$ is cyclic:
Suppose your chosen $anotin A$. Then there must be a sequence in $A$: $a_ndownarrow a$ by definition of $inf$. Now consider the sequence $(a_n-a_m)_{m,n}$ to get a contradiction. (A convergent sequence is Cauchy)
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
$endgroup$
add a comment |
$begingroup$
Observe: $A$ is a group with addition: Assume it is nonempty so $exists xin A$.
i) $x-x=0 in A$
ii) If $xin A$ then $0-x in A$
iii) If $x,yin A$ then $x+y=x-(-y)in A$
Now the statement is basically saying discrete subgroups of $mathbb{R}$ is cyclic:
Suppose your chosen $anotin A$. Then there must be a sequence in $A$: $a_ndownarrow a$ by definition of $inf$. Now consider the sequence $(a_n-a_m)_{m,n}$ to get a contradiction. (A convergent sequence is Cauchy)
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
$endgroup$
Observe: $A$ is a group with addition: Assume it is nonempty so $exists xin A$.
i) $x-x=0 in A$
ii) If $xin A$ then $0-x in A$
iii) If $x,yin A$ then $x+y=x-(-y)in A$
Now the statement is basically saying discrete subgroups of $mathbb{R}$ is cyclic:
Suppose your chosen $anotin A$. Then there must be a sequence in $A$: $a_ndownarrow a$ by definition of $inf$. Now consider the sequence $(a_n-a_m)_{m,n}$ to get a contradiction. (A convergent sequence is Cauchy)
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
edited Dec 10 '18 at 15:42
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
answered Dec 10 '18 at 15:25
user25959user25959
1,573916
1,573916
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%2f3034037%2fshow-that-if-infa-a0-then-a-in-a-and-a-zaz-in-mathbbz%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
