Mixing
Topology Open Balls and Proving Open Sets
$begingroup$
Let $ x ∈ ℝ^n $ and $ ε > 0 $. Prove that the open ball $ B_ε(x) $ is an open set (rigorously!).
I am having trouble with the problem above. If I am understanding it correctly I should be able to set $ x ∈ ℝ^n $ and $ ε > 0 $. Then the open ball $ B_ε(x) $ centered at $ x $ of radius $ ε $ is the set {$ x ∈ ℝ^n | d(x,x_o) < ε $}. If I am correct thus far then a subset $ usubseteqℝ^n $ is open if $ forall$ $ x in u$, $ exists$ $ ε>0 $ such that $ B_ε (x)subseteq u $.
This is as far as I can get. I do not know if this is all I have to do or if I need to do something else? If someone could Help out I would really appreciate it!
general-topology
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
$endgroup$
add a comment |
$begingroup$
Let $ x ∈ ℝ^n $ and $ ε > 0 $. Prove that the open ball $ B_ε(x) $ is an open set (rigorously!).
I am having trouble with the problem above. If I am understanding it correctly I should be able to set $ x ∈ ℝ^n $ and $ ε > 0 $. Then the open ball $ B_ε(x) $ centered at $ x $ of radius $ ε $ is the set {$ x ∈ ℝ^n | d(x,x_o) < ε $}. If I am correct thus far then a subset $ usubseteqℝ^n $ is open if $ forall$ $ x in u$, $ exists$ $ ε>0 $ such that $ B_ε (x)subseteq u $.
This is as far as I can get. I do not know if this is all I have to do or if I need to do something else? If someone could Help out I would really appreciate it!
general-topology
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
$endgroup$
$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29
add a comment |
$begingroup$
Let $ x ∈ ℝ^n $ and $ ε > 0 $. Prove that the open ball $ B_ε(x) $ is an open set (rigorously!).
I am having trouble with the problem above. If I am understanding it correctly I should be able to set $ x ∈ ℝ^n $ and $ ε > 0 $. Then the open ball $ B_ε(x) $ centered at $ x $ of radius $ ε $ is the set {$ x ∈ ℝ^n | d(x,x_o) < ε $}. If I am correct thus far then a subset $ usubseteqℝ^n $ is open if $ forall$ $ x in u$, $ exists$ $ ε>0 $ such that $ B_ε (x)subseteq u $.
This is as far as I can get. I do not know if this is all I have to do or if I need to do something else? If someone could Help out I would really appreciate it!
general-topology
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
$endgroup$
Let $ x ∈ ℝ^n $ and $ ε > 0 $. Prove that the open ball $ B_ε(x) $ is an open set (rigorously!).
I am having trouble with the problem above. If I am understanding it correctly I should be able to set $ x ∈ ℝ^n $ and $ ε > 0 $. Then the open ball $ B_ε(x) $ centered at $ x $ of radius $ ε $ is the set {$ x ∈ ℝ^n | d(x,x_o) < ε $}. If I am correct thus far then a subset $ usubseteqℝ^n $ is open if $ forall$ $ x in u$, $ exists$ $ ε>0 $ such that $ B_ε (x)subseteq u $.
This is as far as I can get. I do not know if this is all I have to do or if I need to do something else? If someone could Help out I would really appreciate it!
general-topology
general-topology
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
edited Jan 31 at 20:54
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Jan 31 at 20:25
DataD96DataD96
295
asked Jan 31 at 20:25
DataD96DataD96
295
asked Jan 31 at 20:25
DataD96DataD96
295
295
$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29
add a comment |
$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29
$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29
$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29
add a comment |
1 Answer
1
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$begingroup$
Let $yin B_epsilon(x)$. Then $d(y,x)=r<epsilon$. Now take any $tin (0,epsilon-r)$. Can you show that $B_t(y)subseteq B_epsilon(x)$?
answered Jan 31 at 20:30
MarkMark
10.5k1622
$endgroup$
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
Let $yin B_epsilon(x)$. Then $d(y,x)=r<epsilon$. Now take any $tin (0,epsilon-r)$. Can you show that $B_t(y)subseteq B_epsilon(x)$?
answered Jan 31 at 20:30
MarkMark
10.5k1622
$endgroup$
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
add a comment |
$begingroup$
Let $yin B_epsilon(x)$. Then $d(y,x)=r<epsilon$. Now take any $tin (0,epsilon-r)$. Can you show that $B_t(y)subseteq B_epsilon(x)$?
answered Jan 31 at 20:30
MarkMark
10.5k1622
$endgroup$
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
add a comment |
$begingroup$
Let $yin B_epsilon(x)$. Then $d(y,x)=r<epsilon$. Now take any $tin (0,epsilon-r)$. Can you show that $B_t(y)subseteq B_epsilon(x)$?
answered Jan 31 at 20:30
MarkMark
10.5k1622
$endgroup$
Let $yin B_epsilon(x)$. Then $d(y,x)=r<epsilon$. Now take any $tin (0,epsilon-r)$. Can you show that $B_t(y)subseteq B_epsilon(x)$?
answered Jan 31 at 20:30
MarkMark
10.5k1622
answered Jan 31 at 20:30
MarkMark
10.5k1622
answered Jan 31 at 20:30
MarkMark
10.5k1622
answered Jan 31 at 20:30
MarkMark
10.5k1622
10.5k1622
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
add a comment |
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
would it be $ d(y,x) $ $ le $ $ d(y,x_o) + d(x_o,x) $ $ lt $ $ t + d(x_0,x) = epsilon $ which would show that $ y in B_epsilon (x) $ which would also show that $ B_t (y) subseteq B_epsilon (x) $ ?
$endgroup$
– DataD96
Jan 31 at 20:47
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
To the proposer: Yes. Exactly right.
$endgroup$
– DanielWainfleet
Jan 31 at 20:52
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
No, $d(y,x)<epsilon$ is something you already know. You need to show that any element in $B_t(y)$ is in $B_epsilon(x)$. So let $zin B_t(y)$. Then $d(z,x)leq d(z,y)+d(y,x)<t+r<(epsilon-r)+r=epsilon$. Hence $z in B_epsilon(x)$.
$endgroup$
– Mark
Jan 31 at 21:20
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
$begingroup$
Oh, okay that makes sense! Thank you for your help!
$endgroup$
– DataD96
Jan 31 at 21:34
add a comment |
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$begingroup$
General idea: given a point in the open ball, $p in B_epsilon(x)$, can you find $delta$ such that $B_delta(p) subset B_epsilon(x)$ ?
$endgroup$
– Joe
Jan 31 at 20:29