When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball?
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I've just learned a throrem which states that : A metric space has the structure of a topological space in which the open sets are unions of balls .
But the theorem only told me there "exist" one topology with respect to the metric.When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball ?
general-topology
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
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add a comment |
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I've just learned a throrem which states that : A metric space has the structure of a topological space in which the open sets are unions of balls .
But the theorem only told me there "exist" one topology with respect to the metric.When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball ?
general-topology
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
$endgroup$
add a comment |
$begingroup$
I've just learned a throrem which states that : A metric space has the structure of a topological space in which the open sets are unions of balls .
But the theorem only told me there "exist" one topology with respect to the metric.When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball ?
general-topology
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
$endgroup$
I've just learned a throrem which states that : A metric space has the structure of a topological space in which the open sets are unions of balls .
But the theorem only told me there "exist" one topology with respect to the metric.When we refer to topology on a metric space $S$ , do we mean the topology generated by open ball ?
general-topology
general-topology
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
asked Dec 28 '18 at 3:53
J.GuoJ.Guo
4029
4029
add a comment |
add a comment |
2 Answers
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Yes, a metric space $(S, d)$ induces a topology $mathcal{T}$ which is generated by the basis $$mathcal{B} = {B(x, r) | x in S text{ and } r > 0}$$
When authors refer to the topology on this metric space $(S, d)$, they usually mean the topology $mathcal{T}$ above, which you can think of as the topology generated by open balls.
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
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Yes. So you can say that a set $U$ is open iff for each $xin U$, there exists an open ball $B(x,r)$ centered at $x$ with $B(x,r)subset U$.
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
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2 Answers
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2 Answers
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Yes, a metric space $(S, d)$ induces a topology $mathcal{T}$ which is generated by the basis $$mathcal{B} = {B(x, r) | x in S text{ and } r > 0}$$
When authors refer to the topology on this metric space $(S, d)$, they usually mean the topology $mathcal{T}$ above, which you can think of as the topology generated by open balls.
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
$endgroup$
add a comment |
$begingroup$
Yes, a metric space $(S, d)$ induces a topology $mathcal{T}$ which is generated by the basis $$mathcal{B} = {B(x, r) | x in S text{ and } r > 0}$$
When authors refer to the topology on this metric space $(S, d)$, they usually mean the topology $mathcal{T}$ above, which you can think of as the topology generated by open balls.
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
$endgroup$
add a comment |
$begingroup$
Yes, a metric space $(S, d)$ induces a topology $mathcal{T}$ which is generated by the basis $$mathcal{B} = {B(x, r) | x in S text{ and } r > 0}$$
When authors refer to the topology on this metric space $(S, d)$, they usually mean the topology $mathcal{T}$ above, which you can think of as the topology generated by open balls.
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
$endgroup$
Yes, a metric space $(S, d)$ induces a topology $mathcal{T}$ which is generated by the basis $$mathcal{B} = {B(x, r) | x in S text{ and } r > 0}$$
When authors refer to the topology on this metric space $(S, d)$, they usually mean the topology $mathcal{T}$ above, which you can think of as the topology generated by open balls.
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
edited Jan 4 at 6:54
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
answered Dec 28 '18 at 4:00
PerturbativePerturbative
4,38121553
4,38121553
add a comment |
add a comment |
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Yes. So you can say that a set $U$ is open iff for each $xin U$, there exists an open ball $B(x,r)$ centered at $x$ with $B(x,r)subset U$.
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
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add a comment |
$begingroup$
Yes. So you can say that a set $U$ is open iff for each $xin U$, there exists an open ball $B(x,r)$ centered at $x$ with $B(x,r)subset U$.
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
Yes. So you can say that a set $U$ is open iff for each $xin U$, there exists an open ball $B(x,r)$ centered at $x$ with $B(x,r)subset U$.
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
$endgroup$
Yes. So you can say that a set $U$ is open iff for each $xin U$, there exists an open ball $B(x,r)$ centered at $x$ with $B(x,r)subset U$.
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
answered Dec 28 '18 at 4:15
Chris CusterChris Custer
13.8k3827
13.8k3827
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