Does adding a continuous inequality constraint over a compact set lead to another compact set?












1












$begingroup$


so the problem is as follows.
I have the vector space $x=[x_1,x_2,...x_N] subseteq R^N, 0 leq x_{1,2,...,N} leq M$
and I extract from it a subset by adding a constraint of this kind:



$X^1={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, epsilon in R}$



Where both $a(x_1),b(x_2)$ are invertible functions in their argument (which is a component of $x$).



$c(x)=a(x_1)-b(x_2)$ is the difference of two invertible functions (even if they were invertible on a single component of the whole vector $x$, so $c(x)$ is not invertible, but I'd say it is continuous).



Now, I'd dare to say that $X^1$ is a compact set, since, the set ${k in R |0leq kleqepsilon}$ is closed and bounded (and hence compact for Heine–Borel) and the starting subsets $0 leq x_{1,2,...,N} leq M$ were compact. I still miss the criteria to apply here, since $c^{-1}(x)$ does not exist, but this is where I got for now.



Now I wonder, if I apply another constraint to $X^1$ and arrive to $X^2$



$X^2={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, 0leq a(x_2)-b(x_3)leqepsilon, epsilon in R}$



would $X^2$ still be compact?



In linear cases, this seems to be intuitive since a linear constraint would lead to a convex polytope (a halfplane + the bounds, so a bounded convex polytope), and the intersection of convex polytopes is still a convex polytope (or I could think of it as products of compact sets, maybe).



Since the various functions are invertible, I tough of this simple example:
let $a(x_1)=x_1$ and $b(x)=x_2^3$
substituting $x_2$ with $b(x_2)$ would mean transforming the $x_2$ axis by stretching it. In the $x_1 - b(x_2)$ plane I obtain that $X^1$ is a halfplane, and is hence compact. Returning to the $x_1 - x_2$ plane means transforming back the $b(x_2)$ to the $x_2$ axis, "squeezing" it, hence preserving the compactness.



This intuitively is true for any invertible function, but it is not a valid proof of course.



What theorem am I missing? is anything that I wrote incorrect?






general-topology convex-analysis compactness







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you intend what you wrote , that x is a set of N elements of R$^N$?
    $endgroup$
    – William Elliot
    Jan 21 at 2:51










  • $begingroup$
    @William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
    $endgroup$
    – user3149593
    Jan 21 at 7:32










  • $begingroup$
    x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
    $endgroup$
    – William Elliot
    Jan 21 at 9:46
















1












$begingroup$


so the problem is as follows.
I have the vector space $x=[x_1,x_2,...x_N] subseteq R^N, 0 leq x_{1,2,...,N} leq M$
and I extract from it a subset by adding a constraint of this kind:



$X^1={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, epsilon in R}$



Where both $a(x_1),b(x_2)$ are invertible functions in their argument (which is a component of $x$).



$c(x)=a(x_1)-b(x_2)$ is the difference of two invertible functions (even if they were invertible on a single component of the whole vector $x$, so $c(x)$ is not invertible, but I'd say it is continuous).



Now, I'd dare to say that $X^1$ is a compact set, since, the set ${k in R |0leq kleqepsilon}$ is closed and bounded (and hence compact for Heine–Borel) and the starting subsets $0 leq x_{1,2,...,N} leq M$ were compact. I still miss the criteria to apply here, since $c^{-1}(x)$ does not exist, but this is where I got for now.



Now I wonder, if I apply another constraint to $X^1$ and arrive to $X^2$



$X^2={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, 0leq a(x_2)-b(x_3)leqepsilon, epsilon in R}$



would $X^2$ still be compact?



In linear cases, this seems to be intuitive since a linear constraint would lead to a convex polytope (a halfplane + the bounds, so a bounded convex polytope), and the intersection of convex polytopes is still a convex polytope (or I could think of it as products of compact sets, maybe).



Since the various functions are invertible, I tough of this simple example:
let $a(x_1)=x_1$ and $b(x)=x_2^3$
substituting $x_2$ with $b(x_2)$ would mean transforming the $x_2$ axis by stretching it. In the $x_1 - b(x_2)$ plane I obtain that $X^1$ is a halfplane, and is hence compact. Returning to the $x_1 - x_2$ plane means transforming back the $b(x_2)$ to the $x_2$ axis, "squeezing" it, hence preserving the compactness.



This intuitively is true for any invertible function, but it is not a valid proof of course.



What theorem am I missing? is anything that I wrote incorrect?






general-topology convex-analysis compactness







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you intend what you wrote , that x is a set of N elements of R$^N$?
    $endgroup$
    – William Elliot
    Jan 21 at 2:51










  • $begingroup$
    @William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
    $endgroup$
    – user3149593
    Jan 21 at 7:32










  • $begingroup$
    x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
    $endgroup$
    – William Elliot
    Jan 21 at 9:46














1












1








1





$begingroup$


so the problem is as follows.
I have the vector space $x=[x_1,x_2,...x_N] subseteq R^N, 0 leq x_{1,2,...,N} leq M$
and I extract from it a subset by adding a constraint of this kind:



$X^1={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, epsilon in R}$



Where both $a(x_1),b(x_2)$ are invertible functions in their argument (which is a component of $x$).



$c(x)=a(x_1)-b(x_2)$ is the difference of two invertible functions (even if they were invertible on a single component of the whole vector $x$, so $c(x)$ is not invertible, but I'd say it is continuous).



Now, I'd dare to say that $X^1$ is a compact set, since, the set ${k in R |0leq kleqepsilon}$ is closed and bounded (and hence compact for Heine–Borel) and the starting subsets $0 leq x_{1,2,...,N} leq M$ were compact. I still miss the criteria to apply here, since $c^{-1}(x)$ does not exist, but this is where I got for now.



Now I wonder, if I apply another constraint to $X^1$ and arrive to $X^2$



$X^2={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, 0leq a(x_2)-b(x_3)leqepsilon, epsilon in R}$



would $X^2$ still be compact?



In linear cases, this seems to be intuitive since a linear constraint would lead to a convex polytope (a halfplane + the bounds, so a bounded convex polytope), and the intersection of convex polytopes is still a convex polytope (or I could think of it as products of compact sets, maybe).



Since the various functions are invertible, I tough of this simple example:
let $a(x_1)=x_1$ and $b(x)=x_2^3$
substituting $x_2$ with $b(x_2)$ would mean transforming the $x_2$ axis by stretching it. In the $x_1 - b(x_2)$ plane I obtain that $X^1$ is a halfplane, and is hence compact. Returning to the $x_1 - x_2$ plane means transforming back the $b(x_2)$ to the $x_2$ axis, "squeezing" it, hence preserving the compactness.



This intuitively is true for any invertible function, but it is not a valid proof of course.



What theorem am I missing? is anything that I wrote incorrect?






general-topology convex-analysis compactness







share|cite|improve this question











$endgroup$




so the problem is as follows.
I have the vector space $x=[x_1,x_2,...x_N] subseteq R^N, 0 leq x_{1,2,...,N} leq M$
and I extract from it a subset by adding a constraint of this kind:



$X^1={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, epsilon in R}$



Where both $a(x_1),b(x_2)$ are invertible functions in their argument (which is a component of $x$).



$c(x)=a(x_1)-b(x_2)$ is the difference of two invertible functions (even if they were invertible on a single component of the whole vector $x$, so $c(x)$ is not invertible, but I'd say it is continuous).



Now, I'd dare to say that $X^1$ is a compact set, since, the set ${k in R |0leq kleqepsilon}$ is closed and bounded (and hence compact for Heine–Borel) and the starting subsets $0 leq x_{1,2,...,N} leq M$ were compact. I still miss the criteria to apply here, since $c^{-1}(x)$ does not exist, but this is where I got for now.



Now I wonder, if I apply another constraint to $X^1$ and arrive to $X^2$



$X^2={x in R^N | 0leq a(x_1)-b(x_2)leqepsilon, 0leq a(x_2)-b(x_3)leqepsilon, epsilon in R}$



would $X^2$ still be compact?



In linear cases, this seems to be intuitive since a linear constraint would lead to a convex polytope (a halfplane + the bounds, so a bounded convex polytope), and the intersection of convex polytopes is still a convex polytope (or I could think of it as products of compact sets, maybe).



Since the various functions are invertible, I tough of this simple example:
let $a(x_1)=x_1$ and $b(x)=x_2^3$
substituting $x_2$ with $b(x_2)$ would mean transforming the $x_2$ axis by stretching it. In the $x_1 - b(x_2)$ plane I obtain that $X^1$ is a halfplane, and is hence compact. Returning to the $x_1 - x_2$ plane means transforming back the $b(x_2)$ to the $x_2$ axis, "squeezing" it, hence preserving the compactness.



This intuitively is true for any invertible function, but it is not a valid proof of course.



What theorem am I missing? is anything that I wrote incorrect?






general-topology convex-analysis compactness




general-topology convex-analysis compactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 7:25







user3149593

















asked Jan 21 at 0:28









user3149593user3149593

295




295








  • 1




    $begingroup$
    Did you intend what you wrote , that x is a set of N elements of R$^N$?
    $endgroup$
    – William Elliot
    Jan 21 at 2:51










  • $begingroup$
    @William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
    $endgroup$
    – user3149593
    Jan 21 at 7:32










  • $begingroup$
    x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
    $endgroup$
    – William Elliot
    Jan 21 at 9:46














  • 1




    $begingroup$
    Did you intend what you wrote , that x is a set of N elements of R$^N$?
    $endgroup$
    – William Elliot
    Jan 21 at 2:51










  • $begingroup$
    @William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
    $endgroup$
    – user3149593
    Jan 21 at 7:32










  • $begingroup$
    x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
    $endgroup$
    – William Elliot
    Jan 21 at 9:46








1




1




$begingroup$
Did you intend what you wrote , that x is a set of N elements of R$^N$?
$endgroup$
– William Elliot
Jan 21 at 2:51




$begingroup$
Did you intend what you wrote , that x is a set of N elements of R$^N$?
$endgroup$
– William Elliot
Jan 21 at 2:51












$begingroup$
@William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
$endgroup$
– user3149593
Jan 21 at 7:32




$begingroup$
@William Thanks for pointing it out, I used the wrong brackets. What I meant is that x has N real components (and so is in R^$N$). All of its components are bounded by 0 and M, and the continuous functions only have one each as argument.
$endgroup$
– user3149593
Jan 21 at 7:32












$begingroup$
x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
$endgroup$
– William Elliot
Jan 21 at 9:46




$begingroup$
x now is a point in R$^N$ but it is not a subset of R$^N$ as you have written.
$endgroup$
– William Elliot
Jan 21 at 9:46










1 Answer
1






active

oldest

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0












$begingroup$

Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
    $endgroup$
    – user3149593
    Jan 21 at 13:32












  • $begingroup$
    $f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
    $endgroup$
    – user87690
    Jan 21 at 13:39










  • $begingroup$
    thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
    $endgroup$
    – user3149593
    Jan 21 at 18:08











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1 Answer
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1 Answer
1






active

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active

oldest

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active

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0












$begingroup$

Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
    $endgroup$
    – user3149593
    Jan 21 at 13:32












  • $begingroup$
    $f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
    $endgroup$
    – user87690
    Jan 21 at 13:39










  • $begingroup$
    thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
    $endgroup$
    – user3149593
    Jan 21 at 18:08
















0












$begingroup$

Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
    $endgroup$
    – user3149593
    Jan 21 at 13:32












  • $begingroup$
    $f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
    $endgroup$
    – user87690
    Jan 21 at 13:39










  • $begingroup$
    thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
    $endgroup$
    – user3149593
    Jan 21 at 18:08














0












0








0





$begingroup$

Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.






share|cite|improve this answer









$endgroup$



Adding such a constraint preserves compactness since it just the intersection with the preimage of a closed set with respect to a continuous function, i.e. $X^2 = X^1 ∩ f^{-1}[C]$ where $f$ is continuous and $C$ is closed. No invertibility of $f$ is needed.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 8:59









user87690user87690

6,5761825




6,5761825












  • $begingroup$
    it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
    $endgroup$
    – user3149593
    Jan 21 at 13:32












  • $begingroup$
    $f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
    $endgroup$
    – user87690
    Jan 21 at 13:39










  • $begingroup$
    thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
    $endgroup$
    – user3149593
    Jan 21 at 18:08


















  • $begingroup$
    it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
    $endgroup$
    – user3149593
    Jan 21 at 13:32












  • $begingroup$
    $f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
    $endgroup$
    – user87690
    Jan 21 at 13:39










  • $begingroup$
    thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
    $endgroup$
    – user3149593
    Jan 21 at 18:08
















$begingroup$
it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
$endgroup$
– user3149593
Jan 21 at 13:32






$begingroup$
it seems strange to me to write $f^{-1}$ if there is no inverse... also couldn't the pre-image be not compact? what properties does the pre-image of a closed set (under a continuous function) have?
$endgroup$
– user3149593
Jan 21 at 13:32














$begingroup$
$f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
$endgroup$
– user87690
Jan 21 at 13:39




$begingroup$
$f^{-1}[C]$ is denotes the set ${x ∈ X: f(x) ∈ C}$. It exists even when $f^{-1}$ as a function does not. Preimage of a closed set under a continuous function is a closed set. In fact, that condition is equivalent to continuity.
$endgroup$
– user87690
Jan 21 at 13:39












$begingroup$
thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
$endgroup$
– user3149593
Jan 21 at 18:08




$begingroup$
thank you, I was not aware of this theorem proofwiki.org/wiki/… and it works just fine.
$endgroup$
– user3149593
Jan 21 at 18:08


















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