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Changing the distance considered on a metric space changes open sets inside?
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I've just started topology on my grade. It makes a lot of sense to start introducing something that we've already been introduced to before on Calculus I, the notion of metric spaces.
My teacher said in class that if we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider.
"After that, we find the motivation to think that distance is not important at all, what's really important is to find the open sets, so welcome to topology." Quoting her.
So I don't see as clear as her this sentence: "If we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider."
Can someone provide an intuitive explanation or a hint to deduce a formal proof to that?
Thanks for your time.
general-topology metric-spaces
edited 1 hour ago
José Carlos Santos
121k16101185
asked 1 hour ago
Relure
1,975833
add a comment |Â
up vote
3
down vote
favorite
I've just started topology on my grade. It makes a lot of sense to start introducing something that we've already been introduced to before on Calculus I, the notion of metric spaces.
My teacher said in class that if we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider.
"After that, we find the motivation to think that distance is not important at all, what's really important is to find the open sets, so welcome to topology." Quoting her.
So I don't see as clear as her this sentence: "If we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider."
Can someone provide an intuitive explanation or a hint to deduce a formal proof to that?
Thanks for your time.
general-topology metric-spaces
edited 1 hour ago
José Carlos Santos
121k16101185
asked 1 hour ago
Relure
1,975833
Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've just started topology on my grade. It makes a lot of sense to start introducing something that we've already been introduced to before on Calculus I, the notion of metric spaces.
My teacher said in class that if we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider.
"After that, we find the motivation to think that distance is not important at all, what's really important is to find the open sets, so welcome to topology." Quoting her.
So I don't see as clear as her this sentence: "If we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider."
Can someone provide an intuitive explanation or a hint to deduce a formal proof to that?
Thanks for your time.
general-topology metric-spaces
edited 1 hour ago
José Carlos Santos
121k16101185
asked 1 hour ago
Relure
1,975833
I've just started topology on my grade. It makes a lot of sense to start introducing something that we've already been introduced to before on Calculus I, the notion of metric spaces.
My teacher said in class that if we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider.
"After that, we find the motivation to think that distance is not important at all, what's really important is to find the open sets, so welcome to topology." Quoting her.
So I don't see as clear as her this sentence: "If we have a metric space $(X,d)$ and we change the distance on it, $(X,d')$, then an open ball on $(X,d)$ is an open ball on $(X,d')$, and that leads us to say that all open sets on $X$ are independent of what distance we choose to consider."
Can someone provide an intuitive explanation or a hint to deduce a formal proof to that?
Thanks for your time.
general-topology metric-spaces
general-topology metric-spaces
edited 1 hour ago
José Carlos Santos
121k16101185
asked 1 hour ago
Relure
1,975833
edited 1 hour ago
José Carlos Santos
121k16101185
asked 1 hour ago
Relure
1,975833
edited 1 hour ago
José Carlos Santos
121k16101185
edited 1 hour ago
José Carlos Santos
121k16101185
edited 1 hour ago
José Carlos Santos
121k16101185
121k16101185
asked 1 hour ago
Relure
1,975833
asked 1 hour ago
Relure
1,975833
asked 1 hour ago
Relure
1,975833
1,975833
Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago
add a comment |Â
Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago
Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago
Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago
add a comment |Â
3 Answers
3
active
oldest
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up vote
3
down vote
I cannot explain that since it is trivially false. For instance, if we, on $mathbb R$, consider the discrete distance$$d(x,y)=begincases1&text if xneq y\0&text otherwise.endcases$$ and if $d'$ is the usual distance, then $0$ is an open set in $(mathbbR,d)$ (being equal to $B_1(0)$), but not in $(mathbbR,d')$.
answered 1 hour ago
José Carlos Santos
121k16101185
add a comment |Â
up vote
2
down vote
I think what your teacher meant is "if we change $(X,d)$ to $(X,d')$ in such a way that open balls in $(X,d)$ coincide with open balls in $(X,d')$ then open sets coincide, meaning they are independent on the choice of the metric". Otherwise the statement is not true.
An example would be $X$ with at least two points and with trivial metrics. For a given real number $c>0$ put
$$d_c(x,y)=begincasesc& xneq y\ 0&textotherwiseendcases$$
and note that for any two different $cneq c'$ we have $d_cneq d_c'$ but open balls coincide and thus generated topologies coincide as well.
And so from the topological point of view the choice of $c$ is not important at all.
answered 1 hour ago
freakish
8,8331524
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
add a comment |Â
up vote
1
down vote
As José Carlos Santos already proved by counterexample, in that generality the statement is false. However, with a slight addition, you get a true statement:
If you have two different metrics $d$, $d'$ on the same set, such that for every point $p$, every open ball of $d$ around $p$ contains an open ball of $d'$ around $p$ and vice versa, then both generate the same topology.
The proof is simple: An open set in a metric-induced topology is, by definition, a set that contains an open ball around each of its points. But since each open ball of one of the metrics contains an open ball of the other metrics around the same point, that first-metric-open sets also contains an open ball of the other metric around each of its points, and therefore is other-metric-open. The same argument works the other way round.
answered 33 mins ago
celtschk
29k75498
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I cannot explain that since it is trivially false. For instance, if we, on $mathbb R$, consider the discrete distance$$d(x,y)=begincases1&text if xneq y\0&text otherwise.endcases$$ and if $d'$ is the usual distance, then $0$ is an open set in $(mathbbR,d)$ (being equal to $B_1(0)$), but not in $(mathbbR,d')$.
answered 1 hour ago
José Carlos Santos
121k16101185
add a comment |Â
up vote
3
down vote
I cannot explain that since it is trivially false. For instance, if we, on $mathbb R$, consider the discrete distance$$d(x,y)=begincases1&text if xneq y\0&text otherwise.endcases$$ and if $d'$ is the usual distance, then $0$ is an open set in $(mathbbR,d)$ (being equal to $B_1(0)$), but not in $(mathbbR,d')$.
answered 1 hour ago
José Carlos Santos
121k16101185
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I cannot explain that since it is trivially false. For instance, if we, on $mathbb R$, consider the discrete distance$$d(x,y)=begincases1&text if xneq y\0&text otherwise.endcases$$ and if $d'$ is the usual distance, then $0$ is an open set in $(mathbbR,d)$ (being equal to $B_1(0)$), but not in $(mathbbR,d')$.
answered 1 hour ago
José Carlos Santos
121k16101185
I cannot explain that since it is trivially false. For instance, if we, on $mathbb R$, consider the discrete distance$$d(x,y)=begincases1&text if xneq y\0&text otherwise.endcases$$ and if $d'$ is the usual distance, then $0$ is an open set in $(mathbbR,d)$ (being equal to $B_1(0)$), but not in $(mathbbR,d')$.
answered 1 hour ago
José Carlos Santos
121k16101185
answered 1 hour ago
José Carlos Santos
121k16101185
answered 1 hour ago
José Carlos Santos
121k16101185
answered 1 hour ago
José Carlos Santos
121k16101185
121k16101185
add a comment |Â
add a comment |Â
up vote
2
down vote
I think what your teacher meant is "if we change $(X,d)$ to $(X,d')$ in such a way that open balls in $(X,d)$ coincide with open balls in $(X,d')$ then open sets coincide, meaning they are independent on the choice of the metric". Otherwise the statement is not true.
An example would be $X$ with at least two points and with trivial metrics. For a given real number $c>0$ put
$$d_c(x,y)=begincasesc& xneq y\ 0&textotherwiseendcases$$
and note that for any two different $cneq c'$ we have $d_cneq d_c'$ but open balls coincide and thus generated topologies coincide as well.
And so from the topological point of view the choice of $c$ is not important at all.
answered 1 hour ago
freakish
8,8331524
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
add a comment |Â
up vote
2
down vote
I think what your teacher meant is "if we change $(X,d)$ to $(X,d')$ in such a way that open balls in $(X,d)$ coincide with open balls in $(X,d')$ then open sets coincide, meaning they are independent on the choice of the metric". Otherwise the statement is not true.
An example would be $X$ with at least two points and with trivial metrics. For a given real number $c>0$ put
$$d_c(x,y)=begincasesc& xneq y\ 0&textotherwiseendcases$$
and note that for any two different $cneq c'$ we have $d_cneq d_c'$ but open balls coincide and thus generated topologies coincide as well.
And so from the topological point of view the choice of $c$ is not important at all.
answered 1 hour ago
freakish
8,8331524
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I think what your teacher meant is "if we change $(X,d)$ to $(X,d')$ in such a way that open balls in $(X,d)$ coincide with open balls in $(X,d')$ then open sets coincide, meaning they are independent on the choice of the metric". Otherwise the statement is not true.
An example would be $X$ with at least two points and with trivial metrics. For a given real number $c>0$ put
$$d_c(x,y)=begincasesc& xneq y\ 0&textotherwiseendcases$$
and note that for any two different $cneq c'$ we have $d_cneq d_c'$ but open balls coincide and thus generated topologies coincide as well.
And so from the topological point of view the choice of $c$ is not important at all.
answered 1 hour ago
freakish
8,8331524
I think what your teacher meant is "if we change $(X,d)$ to $(X,d')$ in such a way that open balls in $(X,d)$ coincide with open balls in $(X,d')$ then open sets coincide, meaning they are independent on the choice of the metric". Otherwise the statement is not true.
An example would be $X$ with at least two points and with trivial metrics. For a given real number $c>0$ put
$$d_c(x,y)=begincasesc& xneq y\ 0&textotherwiseendcases$$
and note that for any two different $cneq c'$ we have $d_cneq d_c'$ but open balls coincide and thus generated topologies coincide as well.
And so from the topological point of view the choice of $c$ is not important at all.
answered 1 hour ago
freakish
8,8331524
edited 1 hour ago
answered 1 hour ago
freakish
8,8331524
answered 1 hour ago
freakish
8,8331524
answered 1 hour ago
freakish
8,8331524
8,8331524
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
add a comment |Â
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
By "coincide" do you mean that every element $xin X$ that has an open ball $B_d$ centered on $x$ has an open ball $B_d'$ centered on $x$ too right?
– Relure
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
@Relure By coincide I mean that $B_d(x,r)=B_d'(x,r)$.
– freakish
1 hour ago
add a comment |Â
up vote
1
down vote
As José Carlos Santos already proved by counterexample, in that generality the statement is false. However, with a slight addition, you get a true statement:
If you have two different metrics $d$, $d'$ on the same set, such that for every point $p$, every open ball of $d$ around $p$ contains an open ball of $d'$ around $p$ and vice versa, then both generate the same topology.
The proof is simple: An open set in a metric-induced topology is, by definition, a set that contains an open ball around each of its points. But since each open ball of one of the metrics contains an open ball of the other metrics around the same point, that first-metric-open sets also contains an open ball of the other metric around each of its points, and therefore is other-metric-open. The same argument works the other way round.
answered 33 mins ago
celtschk
29k75498
add a comment |Â
up vote
1
down vote
As José Carlos Santos already proved by counterexample, in that generality the statement is false. However, with a slight addition, you get a true statement:
If you have two different metrics $d$, $d'$ on the same set, such that for every point $p$, every open ball of $d$ around $p$ contains an open ball of $d'$ around $p$ and vice versa, then both generate the same topology.
The proof is simple: An open set in a metric-induced topology is, by definition, a set that contains an open ball around each of its points. But since each open ball of one of the metrics contains an open ball of the other metrics around the same point, that first-metric-open sets also contains an open ball of the other metric around each of its points, and therefore is other-metric-open. The same argument works the other way round.
answered 33 mins ago
celtschk
29k75498
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As José Carlos Santos already proved by counterexample, in that generality the statement is false. However, with a slight addition, you get a true statement:
If you have two different metrics $d$, $d'$ on the same set, such that for every point $p$, every open ball of $d$ around $p$ contains an open ball of $d'$ around $p$ and vice versa, then both generate the same topology.
The proof is simple: An open set in a metric-induced topology is, by definition, a set that contains an open ball around each of its points. But since each open ball of one of the metrics contains an open ball of the other metrics around the same point, that first-metric-open sets also contains an open ball of the other metric around each of its points, and therefore is other-metric-open. The same argument works the other way round.
answered 33 mins ago
celtschk
29k75498
As José Carlos Santos already proved by counterexample, in that generality the statement is false. However, with a slight addition, you get a true statement:
If you have two different metrics $d$, $d'$ on the same set, such that for every point $p$, every open ball of $d$ around $p$ contains an open ball of $d'$ around $p$ and vice versa, then both generate the same topology.
The proof is simple: An open set in a metric-induced topology is, by definition, a set that contains an open ball around each of its points. But since each open ball of one of the metrics contains an open ball of the other metrics around the same point, that first-metric-open sets also contains an open ball of the other metric around each of its points, and therefore is other-metric-open. The same argument works the other way round.
answered 33 mins ago
celtschk
29k75498
answered 33 mins ago
celtschk
29k75498
answered 33 mins ago
celtschk
29k75498
answered 33 mins ago
celtschk
29k75498
29k75498
add a comment |Â
add a comment |Â
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Perhaps "change the distance" was intended to be (and perhaps was, by context of the conversation) something like "stretch or contract distances". The stretching/contracting doesn't have to be uniform in the sense that there is a constant $alpha > 0$ such that for each $x,y in X$ we have $d'(x,y) = alpha d(x,y),$ although this might have been the context. See Notions of equivalent metrics for more details.
– Dave L. Renfro
26 mins ago