Mixing
Collection of all open balls, centered at the same point, in a dense subset form a base for the containing set?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I'm not even convinced that this is true, and I'm hoping that someone can help me to see why it is true.
I am attempting to prove the following:
Consider a metric space $F$ and a set $E$ that is dense in $F$. Show that the set of all open balls in E, centered at $e in E$ and with rational radii, say $B(e, r_i)$, are a base for $F$.
To show that $B(e, r_i)$ is a base for $F$, I'd need to show that for any open set $T subset F$ and element $t in T$, there exists a ball $B(e, r_t) in B(e, r_i)$ such that the following is true:
$$t in B(e, r_t) subset T$$
But, I don't even believe that this is true. If all of the balls in $B(e, r_i)$ must be centered at a fixed point $e$, then how can one of them always be contained in an arbitrary, open subset of $F$? The picture below is meant to illustrate my confusion.
real-analysis general-topology metric-spaces
asked 1 hour ago
lthompson
708
add a comment |Â
up vote
3
down vote
favorite
I'm not even convinced that this is true, and I'm hoping that someone can help me to see why it is true.
I am attempting to prove the following:
Consider a metric space $F$ and a set $E$ that is dense in $F$. Show that the set of all open balls in E, centered at $e in E$ and with rational radii, say $B(e, r_i)$, are a base for $F$.
To show that $B(e, r_i)$ is a base for $F$, I'd need to show that for any open set $T subset F$ and element $t in T$, there exists a ball $B(e, r_t) in B(e, r_i)$ such that the following is true:
$$t in B(e, r_t) subset T$$
But, I don't even believe that this is true. If all of the balls in $B(e, r_i)$ must be centered at a fixed point $e$, then how can one of them always be contained in an arbitrary, open subset of $F$? The picture below is meant to illustrate my confusion.
real-analysis general-topology metric-spaces
asked 1 hour ago
lthompson
708
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm not even convinced that this is true, and I'm hoping that someone can help me to see why it is true.
I am attempting to prove the following:
Consider a metric space $F$ and a set $E$ that is dense in $F$. Show that the set of all open balls in E, centered at $e in E$ and with rational radii, say $B(e, r_i)$, are a base for $F$.
To show that $B(e, r_i)$ is a base for $F$, I'd need to show that for any open set $T subset F$ and element $t in T$, there exists a ball $B(e, r_t) in B(e, r_i)$ such that the following is true:
$$t in B(e, r_t) subset T$$
But, I don't even believe that this is true. If all of the balls in $B(e, r_i)$ must be centered at a fixed point $e$, then how can one of them always be contained in an arbitrary, open subset of $F$? The picture below is meant to illustrate my confusion.
real-analysis general-topology metric-spaces
asked 1 hour ago
lthompson
708
I'm not even convinced that this is true, and I'm hoping that someone can help me to see why it is true.
I am attempting to prove the following:
Consider a metric space $F$ and a set $E$ that is dense in $F$. Show that the set of all open balls in E, centered at $e in E$ and with rational radii, say $B(e, r_i)$, are a base for $F$.
To show that $B(e, r_i)$ is a base for $F$, I'd need to show that for any open set $T subset F$ and element $t in T$, there exists a ball $B(e, r_t) in B(e, r_i)$ such that the following is true:
$$t in B(e, r_t) subset T$$
But, I don't even believe that this is true. If all of the balls in $B(e, r_i)$ must be centered at a fixed point $e$, then how can one of them always be contained in an arbitrary, open subset of $F$? The picture below is meant to illustrate my confusion.
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked 1 hour ago
lthompson
708
asked 1 hour ago
lthompson
708
edited 1 hour ago
asked 1 hour ago
lthompson
708
asked 1 hour ago
lthompson
708
asked 1 hour ago
lthompson
708
708
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago
add a comment |Â
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
You are correct! The statement, as you write, is untrue.
Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e in E, p in P$. In particular, $e not in P$. However, any union of open sets, all containing $e$, contains $e$.
Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).
answered 1 hour ago
user156213
57828
add a comment |Â
up vote
2
down vote
Minimal counterexample. Take $E = F = a,b$ equipped with the discrete metric:
$$
d(x,y) = begincases
0 & textif $x = y$ \
1 & textif $x not= y$
endcases
$$
The topology defined by $d$ is the discrete topology. However, if you take only one $e in E$, you only get two open balls, $e$ and $F$ and these two balls do not form a basis of the discrete topology.
That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e in E$".
answered 1 hour ago
J.-E. Pin
17.7k21754
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You are correct! The statement, as you write, is untrue.
Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e in E, p in P$. In particular, $e not in P$. However, any union of open sets, all containing $e$, contains $e$.
Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).
answered 1 hour ago
user156213
57828
add a comment |Â
up vote
2
down vote
You are correct! The statement, as you write, is untrue.
Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e in E, p in P$. In particular, $e not in P$. However, any union of open sets, all containing $e$, contains $e$.
Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).
answered 1 hour ago
user156213
57828
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are correct! The statement, as you write, is untrue.
Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e in E, p in P$. In particular, $e not in P$. However, any union of open sets, all containing $e$, contains $e$.
Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).
answered 1 hour ago
user156213
57828
You are correct! The statement, as you write, is untrue.
Every metric space is Hausdorff. Therefore, for $e$ and any other point $p$, there exist disjoint open sets $E$ and $P$ such that $e in E, p in P$. In particular, $e not in P$. However, any union of open sets, all containing $e$, contains $e$.
Therefore P is not the union of open sets containing $e$, so the balls around $e$ are not a base for the metric space (containing at least 2 points).
answered 1 hour ago
user156213
57828
answered 1 hour ago
user156213
57828
answered 1 hour ago
user156213
57828
answered 1 hour ago
user156213
57828
57828
add a comment |Â
add a comment |Â
up vote
2
down vote
Minimal counterexample. Take $E = F = a,b$ equipped with the discrete metric:
$$
d(x,y) = begincases
0 & textif $x = y$ \
1 & textif $x not= y$
endcases
$$
The topology defined by $d$ is the discrete topology. However, if you take only one $e in E$, you only get two open balls, $e$ and $F$ and these two balls do not form a basis of the discrete topology.
That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e in E$".
answered 1 hour ago
J.-E. Pin
17.7k21754
add a comment |Â
up vote
2
down vote
Minimal counterexample. Take $E = F = a,b$ equipped with the discrete metric:
$$
d(x,y) = begincases
0 & textif $x = y$ \
1 & textif $x not= y$
endcases
$$
The topology defined by $d$ is the discrete topology. However, if you take only one $e in E$, you only get two open balls, $e$ and $F$ and these two balls do not form a basis of the discrete topology.
That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e in E$".
answered 1 hour ago
J.-E. Pin
17.7k21754
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Minimal counterexample. Take $E = F = a,b$ equipped with the discrete metric:
$$
d(x,y) = begincases
0 & textif $x = y$ \
1 & textif $x not= y$
endcases
$$
The topology defined by $d$ is the discrete topology. However, if you take only one $e in E$, you only get two open balls, $e$ and $F$ and these two balls do not form a basis of the discrete topology.
That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e in E$".
answered 1 hour ago
J.-E. Pin
17.7k21754
Minimal counterexample. Take $E = F = a,b$ equipped with the discrete metric:
$$
d(x,y) = begincases
0 & textif $x = y$ \
1 & textif $x not= y$
endcases
$$
The topology defined by $d$ is the discrete topology. However, if you take only one $e in E$, you only get two open balls, $e$ and $F$ and these two balls do not form a basis of the discrete topology.
That being said, I fully subscribe to @drhab comment: I would understand the statement as "for some $e in E$".
answered 1 hour ago
J.-E. Pin
17.7k21754
answered 1 hour ago
J.-E. Pin
17.7k21754
answered 1 hour ago
J.-E. Pin
17.7k21754
answered 1 hour ago
J.-E. Pin
17.7k21754
17.7k21754
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2953527%2fcollection-of-all-open-balls-centered-at-the-same-point-in-a-dense-subset-form%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
"...centered at $ein E$..." Shouldn't that be red as "...centered at some $ein E$..."?
– drhab
1 hour ago
@drhab It seems so, but no. In the question I've got, it says "centered at $e in E$", which I've taken to mean a fixed point $e$.
– lthompson
1 hour ago