Prove that the topological closure of a set is definable if the set is definable
$begingroup$
Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.
My attempt at a solution:
Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.
The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:
- $anotin S$
$forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points
Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.
logic first-order-logic predicate-logic model-theory
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
$endgroup$
|
show 5 more comments
$begingroup$
Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.
My attempt at a solution:
Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.
The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:
- $anotin S$
$forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points
Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.
logic first-order-logic predicate-logic model-theory
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
$endgroup$
1
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
1
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18
|
show 5 more comments
$begingroup$
Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.
My attempt at a solution:
Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.
The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:
- $anotin S$
$forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points
Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.
logic first-order-logic predicate-logic model-theory
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
$endgroup$
Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.
My attempt at a solution:
Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.
The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:
- $anotin S$
$forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points
Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.
logic first-order-logic predicate-logic model-theory
logic first-order-logic predicate-logic model-theory
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
edited Dec 5 '18 at 19:57
Asaf Karagila♦
305k33436766
305k33436766
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
asked Dec 5 '18 at 19:11
quanticboltquanticbolt
769512
769512
1
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
1
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18
|
show 5 more comments
1
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
1
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18
1
1
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
1
1
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:
$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$
You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
$endgroup$
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%2f3027507%2fprove-that-the-topological-closure-of-a-set-is-definable-if-the-set-is-definable%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
$begingroup$
As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:
$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$
You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
$endgroup$
add a comment |
$begingroup$
As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:
$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$
You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
$endgroup$
add a comment |
$begingroup$
As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:
$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$
You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
$endgroup$
As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:
$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$
You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
answered Dec 6 '18 at 1:42
Alex KruckmanAlex Kruckman
27.6k32658
27.6k32658
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%2f3027507%2fprove-that-the-topological-closure-of-a-set-is-definable-if-the-set-is-definable%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
$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54
$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04
$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12
$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15
1
$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18