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For metric space $(X, d)$ and fixed $ain X$, prove function from induced topology to $Bbb R$ defined by $f(x) = d(a, x)$ is continuous.
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Let the induced topology be $(X, tau)$. Let $(x, y) subset Bbb R$. I'm required to prove that $f^-1((x, y))$ is open in $tau$
$f^-1((x, y)) = p in X : x < d(p, a) < y $
My thinking is to show this is a union of open balls. Don't know if this is allowed, but for each $p in f^-1((x, y))$, find a $delta > 0$ such that $B_delta(p) subseteq f^-1((x, y)) $
Am I going in the right direction with this? Not sure how to proceed regardless.
general-topology metric-spaces continuity
asked yesterday
Mattice Verhoeven
567410
add a comment |Â
up vote
1
down vote
favorite
Let the induced topology be $(X, tau)$. Let $(x, y) subset Bbb R$. I'm required to prove that $f^-1((x, y))$ is open in $tau$
$f^-1((x, y)) = p in X : x < d(p, a) < y $
My thinking is to show this is a union of open balls. Don't know if this is allowed, but for each $p in f^-1((x, y))$, find a $delta > 0$ such that $B_delta(p) subseteq f^-1((x, y)) $
Am I going in the right direction with this? Not sure how to proceed regardless.
general-topology metric-spaces continuity
asked yesterday
Mattice Verhoeven
567410
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let the induced topology be $(X, tau)$. Let $(x, y) subset Bbb R$. I'm required to prove that $f^-1((x, y))$ is open in $tau$
$f^-1((x, y)) = p in X : x < d(p, a) < y $
My thinking is to show this is a union of open balls. Don't know if this is allowed, but for each $p in f^-1((x, y))$, find a $delta > 0$ such that $B_delta(p) subseteq f^-1((x, y)) $
Am I going in the right direction with this? Not sure how to proceed regardless.
general-topology metric-spaces continuity
asked yesterday
Mattice Verhoeven
567410
Let the induced topology be $(X, tau)$. Let $(x, y) subset Bbb R$. I'm required to prove that $f^-1((x, y))$ is open in $tau$
$f^-1((x, y)) = p in X : x < d(p, a) < y $
My thinking is to show this is a union of open balls. Don't know if this is allowed, but for each $p in f^-1((x, y))$, find a $delta > 0$ such that $B_delta(p) subseteq f^-1((x, y)) $
Am I going in the right direction with this? Not sure how to proceed regardless.
general-topology metric-spaces continuity
asked yesterday
Mattice Verhoeven
567410
asked yesterday
Mattice Verhoeven
567410
asked yesterday
Mattice Verhoeven
567410
asked yesterday
Mattice Verhoeven
567410
567410
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
It might be easier to prove it directly as in:
$f(x) = d(x,a) le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$
we get $|f(x)-f(y)| le d(x,y)$ from which continuity follows.
Since $f$ is continuous, $f^-1 ((x,y))$ is open.
answered yesterday
copper.hat
122k556155
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
add a comment |Â
up vote
1
down vote
You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^-1((x,y))$ is an element of an open ball that is contained in $f^-1((x,y))$.
Let $p$ be an arbitrary element of $f^-1((x,y))$. Let $r = minleft frac2, frac2 right$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^-1((x,y))$ and hence we conclude that $f^-1((x,y))$ is open.
answered yesterday
Stefan4024
27.6k52874
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It might be easier to prove it directly as in:
$f(x) = d(x,a) le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$
we get $|f(x)-f(y)| le d(x,y)$ from which continuity follows.
Since $f$ is continuous, $f^-1 ((x,y))$ is open.
answered yesterday
copper.hat
122k556155
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
add a comment |Â
up vote
3
down vote
accepted
It might be easier to prove it directly as in:
$f(x) = d(x,a) le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$
we get $|f(x)-f(y)| le d(x,y)$ from which continuity follows.
Since $f$ is continuous, $f^-1 ((x,y))$ is open.
answered yesterday
copper.hat
122k556155
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It might be easier to prove it directly as in:
$f(x) = d(x,a) le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$
we get $|f(x)-f(y)| le d(x,y)$ from which continuity follows.
Since $f$ is continuous, $f^-1 ((x,y))$ is open.
answered yesterday
copper.hat
122k556155
It might be easier to prove it directly as in:
$f(x) = d(x,a) le f(y)+ d(x,y) = f(y) + d(x,y)$. Swapping the roles of $x,y$
we get $|f(x)-f(y)| le d(x,y)$ from which continuity follows.
Since $f$ is continuous, $f^-1 ((x,y))$ is open.
answered yesterday
copper.hat
122k556155
edited yesterday
answered yesterday
copper.hat
122k556155
answered yesterday
copper.hat
122k556155
answered yesterday
copper.hat
122k556155
122k556155
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
add a comment |Â
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
So think we can use that definition of continuity? Being a topology textbook, it has using the topology definition.
– Mattice Verhoeven
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@TheoreticalEconomist: Indeed it is, thanks!
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
@MatticeVerhoeven: The topology is the induced one, so for a given $x$ and $epsilon>0$ you see that if $y in B(x,epsilon)$ then $|f(x)-f(y)| < epsilon$. Note that $B(x,epsilon) = y $.
– copper.hat
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
Sorry I'm confused about what you're trying to do. is $x in Bbb R$ or $x in X$
– Mattice Verhoeven
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
$x in X$. I am just relating the metric definition of open with the topological one.
– copper.hat
yesterday
add a comment |Â
up vote
1
down vote
You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^-1((x,y))$ is an element of an open ball that is contained in $f^-1((x,y))$.
Let $p$ be an arbitrary element of $f^-1((x,y))$. Let $r = minleft frac2, frac2 right$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^-1((x,y))$ and hence we conclude that $f^-1((x,y))$ is open.
answered yesterday
Stefan4024
27.6k52874
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
add a comment |Â
up vote
1
down vote
You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^-1((x,y))$ is an element of an open ball that is contained in $f^-1((x,y))$.
Let $p$ be an arbitrary element of $f^-1((x,y))$. Let $r = minleft frac2, frac2 right$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^-1((x,y))$ and hence we conclude that $f^-1((x,y))$ is open.
answered yesterday
Stefan4024
27.6k52874
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^-1((x,y))$ is an element of an open ball that is contained in $f^-1((x,y))$.
Let $p$ be an arbitrary element of $f^-1((x,y))$. Let $r = minleft frac2, frac2 right$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^-1((x,y))$ and hence we conclude that $f^-1((x,y))$ is open.
answered yesterday
Stefan4024
27.6k52874
You don't have to show that it's a union of open balls (at least not explicitly). You can prove that each element of $f^-1((x,y))$ is an element of an open ball that is contained in $f^-1((x,y))$.
Let $p$ be an arbitrary element of $f^-1((x,y))$. Let $r = minleft frac2, frac2 right$. Then obviously the ball $B_r(a)$ is an open ball contained in $f^-1((x,y))$ and hence we conclude that $f^-1((x,y))$ is open.
answered yesterday
Stefan4024
27.6k52874
edited yesterday
answered yesterday
Stefan4024
27.6k52874
answered yesterday
Stefan4024
27.6k52874
answered yesterday
Stefan4024
27.6k52874
27.6k52874
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
add a comment |Â
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
What does it mean to have $d(x, a)$ where $x in Bbb R, a in X$?
– Mattice Verhoeven
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
@MatticeVerhoeven Sorry it should be $p$ instead of $x$.
– Stefan4024
yesterday
add a comment |Â
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