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Proof Verification: Existence of a path $f: [a,b] to X$ implies the existence of a path $g: [0,1] to X$.
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Can somebody please tell me if this proof is correct? This is not homework. A wise user told me this result in a previous question and I wanted to prove it before using it. It's a pretty trivial proof, essentially boiling down to the continuity of a polynomial (I'm aware that this is a result from calculus but I didn't want to cut corners). I just made a bunch of particularly stupid mistakes in a proof of a claim related to this and I want to ensure I am not missing anything.
Let $k: [0,1] rightarrow [a,b]$ be defined by $k(t) = a + t(b-a)$. Let X be a path connected space, and let $f:[a,b] rightarrow X$ be a path in X from $x_1$ to $x_2$.
Lemma : $k$ is continuous.
Notice that $[0,1]$ and $[a,b]$ are subspaces of the metric space $(mathbbR,d)$, where $d$ is the standard metric for $mathbbR$ (defined by $d(x,y) = |x-y|$). Hence, the restriction $d_1$ of $d$ to $[0,1]$ x $[0,1]$ is a metric for the subspace topology on $[0,1]$ and the restriction $d_2$ of $d$ to $[a,b]$ x $[a,b]$ is a metric for the subspace topology on $[a,b]$ (proof was done earlier, don't worry).
We first want to show that $k$ is order preserving. This is pretty clear, but fix $x,y in [0,1]$ s.t. $x < y$. Then, $x(b-a) < y(b-a)$ because $(b-a) > 0$. Then, $a + x(b-a) < a + y(b-a)$ so that $f(x) < f(y)$.
Fix $x in [0,1]$ and $epsilon > 0$ (and forget that we previously also fixed y in that interval).
Notice that if $y in B_d_1(x,fracepsilonb-a)$, then $d_1(x,y) = d_1(y,x) = |y-x| <fracepsilonb-a$. Then, $x - fracepsilonb-a < y < x + fracepsilonb-a$. Notice that $k(x) - epsilon = a - epsilon + x(b-a) = a + (x - fracepsilonb-a)(b-a) = k(x - fracepsilonb-a)$ and $k(x) + epsilon = a + epsilon + x(b-a) = a + (x + fracepsilonb-a)(b-a) = k(x + fracepsilonb-a)$. Then, $k(x) - epsilon < f(y) < k(x) + epsilon$ because $f$ is order preserving. That is, $|k(y)-k(x)| = d_2(k(y),k(x)) = d_2(k(x),k(y)) < epsilon$.
We have shown that if $d_1(x,y) < fracepsilonb-a$, then $d_2(f(x),f(y)) < epsilon$. Then, $k$ is continuous by Theorem 21.1. Q.E.D
Claim: If $f: [a,b] rightarrow X$ is a path in X from $x_1$ to $x_2$, then $f':[0,1] rightarrow X$ defined by $f'(t) = (f circ k)(t)$ is also a path in X from $x_1$ to $x_2$.
It is clear that $f'(0) = x_1$ and that $f'(1) = x_2$. Notice that $f'$ is continuous by Theorem 18.2(c) because $k$ is continuous by the lemma and $f$ is continuous by hypothesis. We conclude that $f'$ is a path in X. Q.E.D.
Theorems:
Theorem 21.1: Let $f: X rightarrow Y$; let X and Y be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x in X$ and given $epsilon > 0$, there exists $delta > 0$ s.t. $$d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon.$$
Theorem 18.2(c): Let X,Y,Z be topological spaces. If $f: X rightarrow Y$ and $g:Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.
If anybody out there notices any mistakes, please tell me (It's getting late so I might have missed something). Any comments or criticisms are also very welcome. Thank you! :)
general-topology path-connected
edited Aug 6 at 6:41
Sou
2,7062820
asked Aug 6 at 6:05
TenaxPropositi
84
add a comment |Â
up vote
0
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Can somebody please tell me if this proof is correct? This is not homework. A wise user told me this result in a previous question and I wanted to prove it before using it. It's a pretty trivial proof, essentially boiling down to the continuity of a polynomial (I'm aware that this is a result from calculus but I didn't want to cut corners). I just made a bunch of particularly stupid mistakes in a proof of a claim related to this and I want to ensure I am not missing anything.
Let $k: [0,1] rightarrow [a,b]$ be defined by $k(t) = a + t(b-a)$. Let X be a path connected space, and let $f:[a,b] rightarrow X$ be a path in X from $x_1$ to $x_2$.
Lemma : $k$ is continuous.
Notice that $[0,1]$ and $[a,b]$ are subspaces of the metric space $(mathbbR,d)$, where $d$ is the standard metric for $mathbbR$ (defined by $d(x,y) = |x-y|$). Hence, the restriction $d_1$ of $d$ to $[0,1]$ x $[0,1]$ is a metric for the subspace topology on $[0,1]$ and the restriction $d_2$ of $d$ to $[a,b]$ x $[a,b]$ is a metric for the subspace topology on $[a,b]$ (proof was done earlier, don't worry).
We first want to show that $k$ is order preserving. This is pretty clear, but fix $x,y in [0,1]$ s.t. $x < y$. Then, $x(b-a) < y(b-a)$ because $(b-a) > 0$. Then, $a + x(b-a) < a + y(b-a)$ so that $f(x) < f(y)$.
Fix $x in [0,1]$ and $epsilon > 0$ (and forget that we previously also fixed y in that interval).
Notice that if $y in B_d_1(x,fracepsilonb-a)$, then $d_1(x,y) = d_1(y,x) = |y-x| <fracepsilonb-a$. Then, $x - fracepsilonb-a < y < x + fracepsilonb-a$. Notice that $k(x) - epsilon = a - epsilon + x(b-a) = a + (x - fracepsilonb-a)(b-a) = k(x - fracepsilonb-a)$ and $k(x) + epsilon = a + epsilon + x(b-a) = a + (x + fracepsilonb-a)(b-a) = k(x + fracepsilonb-a)$. Then, $k(x) - epsilon < f(y) < k(x) + epsilon$ because $f$ is order preserving. That is, $|k(y)-k(x)| = d_2(k(y),k(x)) = d_2(k(x),k(y)) < epsilon$.
We have shown that if $d_1(x,y) < fracepsilonb-a$, then $d_2(f(x),f(y)) < epsilon$. Then, $k$ is continuous by Theorem 21.1. Q.E.D
Claim: If $f: [a,b] rightarrow X$ is a path in X from $x_1$ to $x_2$, then $f':[0,1] rightarrow X$ defined by $f'(t) = (f circ k)(t)$ is also a path in X from $x_1$ to $x_2$.
It is clear that $f'(0) = x_1$ and that $f'(1) = x_2$. Notice that $f'$ is continuous by Theorem 18.2(c) because $k$ is continuous by the lemma and $f$ is continuous by hypothesis. We conclude that $f'$ is a path in X. Q.E.D.
Theorems:
Theorem 21.1: Let $f: X rightarrow Y$; let X and Y be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x in X$ and given $epsilon > 0$, there exists $delta > 0$ s.t. $$d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon.$$
Theorem 18.2(c): Let X,Y,Z be topological spaces. If $f: X rightarrow Y$ and $g:Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.
If anybody out there notices any mistakes, please tell me (It's getting late so I might have missed something). Any comments or criticisms are also very welcome. Thank you! :)
general-topology path-connected
edited Aug 6 at 6:41
Sou
2,7062820
asked Aug 6 at 6:05
TenaxPropositi
84
3
It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can somebody please tell me if this proof is correct? This is not homework. A wise user told me this result in a previous question and I wanted to prove it before using it. It's a pretty trivial proof, essentially boiling down to the continuity of a polynomial (I'm aware that this is a result from calculus but I didn't want to cut corners). I just made a bunch of particularly stupid mistakes in a proof of a claim related to this and I want to ensure I am not missing anything.
Let $k: [0,1] rightarrow [a,b]$ be defined by $k(t) = a + t(b-a)$. Let X be a path connected space, and let $f:[a,b] rightarrow X$ be a path in X from $x_1$ to $x_2$.
Lemma : $k$ is continuous.
Notice that $[0,1]$ and $[a,b]$ are subspaces of the metric space $(mathbbR,d)$, where $d$ is the standard metric for $mathbbR$ (defined by $d(x,y) = |x-y|$). Hence, the restriction $d_1$ of $d$ to $[0,1]$ x $[0,1]$ is a metric for the subspace topology on $[0,1]$ and the restriction $d_2$ of $d$ to $[a,b]$ x $[a,b]$ is a metric for the subspace topology on $[a,b]$ (proof was done earlier, don't worry).
We first want to show that $k$ is order preserving. This is pretty clear, but fix $x,y in [0,1]$ s.t. $x < y$. Then, $x(b-a) < y(b-a)$ because $(b-a) > 0$. Then, $a + x(b-a) < a + y(b-a)$ so that $f(x) < f(y)$.
Fix $x in [0,1]$ and $epsilon > 0$ (and forget that we previously also fixed y in that interval).
Notice that if $y in B_d_1(x,fracepsilonb-a)$, then $d_1(x,y) = d_1(y,x) = |y-x| <fracepsilonb-a$. Then, $x - fracepsilonb-a < y < x + fracepsilonb-a$. Notice that $k(x) - epsilon = a - epsilon + x(b-a) = a + (x - fracepsilonb-a)(b-a) = k(x - fracepsilonb-a)$ and $k(x) + epsilon = a + epsilon + x(b-a) = a + (x + fracepsilonb-a)(b-a) = k(x + fracepsilonb-a)$. Then, $k(x) - epsilon < f(y) < k(x) + epsilon$ because $f$ is order preserving. That is, $|k(y)-k(x)| = d_2(k(y),k(x)) = d_2(k(x),k(y)) < epsilon$.
We have shown that if $d_1(x,y) < fracepsilonb-a$, then $d_2(f(x),f(y)) < epsilon$. Then, $k$ is continuous by Theorem 21.1. Q.E.D
Claim: If $f: [a,b] rightarrow X$ is a path in X from $x_1$ to $x_2$, then $f':[0,1] rightarrow X$ defined by $f'(t) = (f circ k)(t)$ is also a path in X from $x_1$ to $x_2$.
It is clear that $f'(0) = x_1$ and that $f'(1) = x_2$. Notice that $f'$ is continuous by Theorem 18.2(c) because $k$ is continuous by the lemma and $f$ is continuous by hypothesis. We conclude that $f'$ is a path in X. Q.E.D.
Theorems:
Theorem 21.1: Let $f: X rightarrow Y$; let X and Y be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x in X$ and given $epsilon > 0$, there exists $delta > 0$ s.t. $$d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon.$$
Theorem 18.2(c): Let X,Y,Z be topological spaces. If $f: X rightarrow Y$ and $g:Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.
If anybody out there notices any mistakes, please tell me (It's getting late so I might have missed something). Any comments or criticisms are also very welcome. Thank you! :)
general-topology path-connected
edited Aug 6 at 6:41
Sou
2,7062820
asked Aug 6 at 6:05
TenaxPropositi
84
Can somebody please tell me if this proof is correct? This is not homework. A wise user told me this result in a previous question and I wanted to prove it before using it. It's a pretty trivial proof, essentially boiling down to the continuity of a polynomial (I'm aware that this is a result from calculus but I didn't want to cut corners). I just made a bunch of particularly stupid mistakes in a proof of a claim related to this and I want to ensure I am not missing anything.
Let $k: [0,1] rightarrow [a,b]$ be defined by $k(t) = a + t(b-a)$. Let X be a path connected space, and let $f:[a,b] rightarrow X$ be a path in X from $x_1$ to $x_2$.
Lemma : $k$ is continuous.
Notice that $[0,1]$ and $[a,b]$ are subspaces of the metric space $(mathbbR,d)$, where $d$ is the standard metric for $mathbbR$ (defined by $d(x,y) = |x-y|$). Hence, the restriction $d_1$ of $d$ to $[0,1]$ x $[0,1]$ is a metric for the subspace topology on $[0,1]$ and the restriction $d_2$ of $d$ to $[a,b]$ x $[a,b]$ is a metric for the subspace topology on $[a,b]$ (proof was done earlier, don't worry).
We first want to show that $k$ is order preserving. This is pretty clear, but fix $x,y in [0,1]$ s.t. $x < y$. Then, $x(b-a) < y(b-a)$ because $(b-a) > 0$. Then, $a + x(b-a) < a + y(b-a)$ so that $f(x) < f(y)$.
Fix $x in [0,1]$ and $epsilon > 0$ (and forget that we previously also fixed y in that interval).
Notice that if $y in B_d_1(x,fracepsilonb-a)$, then $d_1(x,y) = d_1(y,x) = |y-x| <fracepsilonb-a$. Then, $x - fracepsilonb-a < y < x + fracepsilonb-a$. Notice that $k(x) - epsilon = a - epsilon + x(b-a) = a + (x - fracepsilonb-a)(b-a) = k(x - fracepsilonb-a)$ and $k(x) + epsilon = a + epsilon + x(b-a) = a + (x + fracepsilonb-a)(b-a) = k(x + fracepsilonb-a)$. Then, $k(x) - epsilon < f(y) < k(x) + epsilon$ because $f$ is order preserving. That is, $|k(y)-k(x)| = d_2(k(y),k(x)) = d_2(k(x),k(y)) < epsilon$.
We have shown that if $d_1(x,y) < fracepsilonb-a$, then $d_2(f(x),f(y)) < epsilon$. Then, $k$ is continuous by Theorem 21.1. Q.E.D
Claim: If $f: [a,b] rightarrow X$ is a path in X from $x_1$ to $x_2$, then $f':[0,1] rightarrow X$ defined by $f'(t) = (f circ k)(t)$ is also a path in X from $x_1$ to $x_2$.
It is clear that $f'(0) = x_1$ and that $f'(1) = x_2$. Notice that $f'$ is continuous by Theorem 18.2(c) because $k$ is continuous by the lemma and $f$ is continuous by hypothesis. We conclude that $f'$ is a path in X. Q.E.D.
Theorems:
Theorem 21.1: Let $f: X rightarrow Y$; let X and Y be metrizable with metrics $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x in X$ and given $epsilon > 0$, there exists $delta > 0$ s.t. $$d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon.$$
Theorem 18.2(c): Let X,Y,Z be topological spaces. If $f: X rightarrow Y$ and $g:Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.
If anybody out there notices any mistakes, please tell me (It's getting late so I might have missed something). Any comments or criticisms are also very welcome. Thank you! :)
general-topology path-connected
edited Aug 6 at 6:41
Sou
2,7062820
asked Aug 6 at 6:05
TenaxPropositi
84
edited Aug 6 at 6:41
Sou
2,7062820
edited Aug 6 at 6:41
Sou
2,7062820
edited Aug 6 at 6:41
Sou
2,7062820
2,7062820
asked Aug 6 at 6:05
TenaxPropositi
84
asked Aug 6 at 6:05
TenaxPropositi
84
asked Aug 6 at 6:05
TenaxPropositi
84
84
3
It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42
add a comment |Â
3
It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42
3
3
It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42
It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42
add a comment |Â
1 Answer
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Anytime you see something like this, make use of the fact that $[0,1]$ and $[a,b]$ are homeomorphic. Therefore, maps out of them should act the same way, which is what your problem is getting at.
As the commenter points out, the map $k: [0,1] to [a,b]$ defined by
$$
k(t) = a + t(b-a)
$$
is certainly continuous as it is linear. (It's also a homeomorphism, but you don't need that for your exact question. I'll come back to this.) Then given any path $f: [a,b] to X$ we have the path $g = f circ k: [0,1] to X$ as desired, since compositions of continuous maps are still continuous. (And you said all this.)
Bonus fact: since $k$ is a homeo, the converse is true by the same argument. The inverse to $k$ is $h: [a,b] to [0,1]$ defined by
$$
h(t) = fract-ab-a
$$
(again linear, again continuous). So the exact same argument shows that every path out of $[0,1]$ gives a path out of $[a,b]$. In fact, these two "translations of domain" give a bijection between the sets of paths out of these two different (but homeomorphic) domains. This nice symmetry is a good justification of why the definition of a homeomorphism is not as a continuous bijection, but as a continuous bijection with continuous inverse.
answered Aug 6 at 14:17
Randall
7,2471825
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Anytime you see something like this, make use of the fact that $[0,1]$ and $[a,b]$ are homeomorphic. Therefore, maps out of them should act the same way, which is what your problem is getting at.
As the commenter points out, the map $k: [0,1] to [a,b]$ defined by
$$
k(t) = a + t(b-a)
$$
is certainly continuous as it is linear. (It's also a homeomorphism, but you don't need that for your exact question. I'll come back to this.) Then given any path $f: [a,b] to X$ we have the path $g = f circ k: [0,1] to X$ as desired, since compositions of continuous maps are still continuous. (And you said all this.)
Bonus fact: since $k$ is a homeo, the converse is true by the same argument. The inverse to $k$ is $h: [a,b] to [0,1]$ defined by
$$
h(t) = fract-ab-a
$$
(again linear, again continuous). So the exact same argument shows that every path out of $[0,1]$ gives a path out of $[a,b]$. In fact, these two "translations of domain" give a bijection between the sets of paths out of these two different (but homeomorphic) domains. This nice symmetry is a good justification of why the definition of a homeomorphism is not as a continuous bijection, but as a continuous bijection with continuous inverse.
answered Aug 6 at 14:17
Randall
7,2471825
add a comment |Â
up vote
0
down vote
Anytime you see something like this, make use of the fact that $[0,1]$ and $[a,b]$ are homeomorphic. Therefore, maps out of them should act the same way, which is what your problem is getting at.
As the commenter points out, the map $k: [0,1] to [a,b]$ defined by
$$
k(t) = a + t(b-a)
$$
is certainly continuous as it is linear. (It's also a homeomorphism, but you don't need that for your exact question. I'll come back to this.) Then given any path $f: [a,b] to X$ we have the path $g = f circ k: [0,1] to X$ as desired, since compositions of continuous maps are still continuous. (And you said all this.)
Bonus fact: since $k$ is a homeo, the converse is true by the same argument. The inverse to $k$ is $h: [a,b] to [0,1]$ defined by
$$
h(t) = fract-ab-a
$$
(again linear, again continuous). So the exact same argument shows that every path out of $[0,1]$ gives a path out of $[a,b]$. In fact, these two "translations of domain" give a bijection between the sets of paths out of these two different (but homeomorphic) domains. This nice symmetry is a good justification of why the definition of a homeomorphism is not as a continuous bijection, but as a continuous bijection with continuous inverse.
answered Aug 6 at 14:17
Randall
7,2471825
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Anytime you see something like this, make use of the fact that $[0,1]$ and $[a,b]$ are homeomorphic. Therefore, maps out of them should act the same way, which is what your problem is getting at.
As the commenter points out, the map $k: [0,1] to [a,b]$ defined by
$$
k(t) = a + t(b-a)
$$
is certainly continuous as it is linear. (It's also a homeomorphism, but you don't need that for your exact question. I'll come back to this.) Then given any path $f: [a,b] to X$ we have the path $g = f circ k: [0,1] to X$ as desired, since compositions of continuous maps are still continuous. (And you said all this.)
Bonus fact: since $k$ is a homeo, the converse is true by the same argument. The inverse to $k$ is $h: [a,b] to [0,1]$ defined by
$$
h(t) = fract-ab-a
$$
(again linear, again continuous). So the exact same argument shows that every path out of $[0,1]$ gives a path out of $[a,b]$. In fact, these two "translations of domain" give a bijection between the sets of paths out of these two different (but homeomorphic) domains. This nice symmetry is a good justification of why the definition of a homeomorphism is not as a continuous bijection, but as a continuous bijection with continuous inverse.
answered Aug 6 at 14:17
Randall
7,2471825
Anytime you see something like this, make use of the fact that $[0,1]$ and $[a,b]$ are homeomorphic. Therefore, maps out of them should act the same way, which is what your problem is getting at.
As the commenter points out, the map $k: [0,1] to [a,b]$ defined by
$$
k(t) = a + t(b-a)
$$
is certainly continuous as it is linear. (It's also a homeomorphism, but you don't need that for your exact question. I'll come back to this.) Then given any path $f: [a,b] to X$ we have the path $g = f circ k: [0,1] to X$ as desired, since compositions of continuous maps are still continuous. (And you said all this.)
Bonus fact: since $k$ is a homeo, the converse is true by the same argument. The inverse to $k$ is $h: [a,b] to [0,1]$ defined by
$$
h(t) = fract-ab-a
$$
(again linear, again continuous). So the exact same argument shows that every path out of $[0,1]$ gives a path out of $[a,b]$. In fact, these two "translations of domain" give a bijection between the sets of paths out of these two different (but homeomorphic) domains. This nice symmetry is a good justification of why the definition of a homeomorphism is not as a continuous bijection, but as a continuous bijection with continuous inverse.
answered Aug 6 at 14:17
Randall
7,2471825
answered Aug 6 at 14:17
Randall
7,2471825
answered Aug 6 at 14:17
Randall
7,2471825
answered Aug 6 at 14:17
Randall
7,2471825
7,2471825
add a comment |Â
add a comment |Â
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It's easier if you argue that $k(t) = a + t(b-a)$ is continous because it's just a linear map and $f circ k$ is continous since it's a composition of continous functions.
– Sou
Aug 6 at 6:42