Linear continuous bijection but not open.
$begingroup$
I have the next question.
Let $l^1$ be the set of sequences $(a_1,a_2,ldots, )$ such that $sum |a_k|<infty$. If we consider norm $|.|_1$ and the supremum norm $|.|_{s}$, then $(l^1,|.|_1)$ is complete , while $(l^1,|.|_s)$ is not complete.
Let $id:(l^1,|.|_1)to (l^1,|.|_s), xto x$ is a continuous bijection but is not open.
Why is not open?
real-analysis functional-analysis open-map
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
I have the next question.
Let $l^1$ be the set of sequences $(a_1,a_2,ldots, )$ such that $sum |a_k|<infty$. If we consider norm $|.|_1$ and the supremum norm $|.|_{s}$, then $(l^1,|.|_1)$ is complete , while $(l^1,|.|_s)$ is not complete.
Let $id:(l^1,|.|_1)to (l^1,|.|_s), xto x$ is a continuous bijection but is not open.
Why is not open?
real-analysis functional-analysis open-map
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
$endgroup$
add a comment |
$begingroup$
I have the next question.
Let $l^1$ be the set of sequences $(a_1,a_2,ldots, )$ such that $sum |a_k|<infty$. If we consider norm $|.|_1$ and the supremum norm $|.|_{s}$, then $(l^1,|.|_1)$ is complete , while $(l^1,|.|_s)$ is not complete.
Let $id:(l^1,|.|_1)to (l^1,|.|_s), xto x$ is a continuous bijection but is not open.
Why is not open?
real-analysis functional-analysis open-map
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
$endgroup$
I have the next question.
Let $l^1$ be the set of sequences $(a_1,a_2,ldots, )$ such that $sum |a_k|<infty$. If we consider norm $|.|_1$ and the supremum norm $|.|_{s}$, then $(l^1,|.|_1)$ is complete , while $(l^1,|.|_s)$ is not complete.
Let $id:(l^1,|.|_1)to (l^1,|.|_s), xto x$ is a continuous bijection but is not open.
Why is not open?
real-analysis functional-analysis open-map
real-analysis functional-analysis open-map
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
asked Dec 19 '18 at 15:59
eraldcoileraldcoil
395211
395211
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $vert xvert_1 le C vert xvert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,dots,1,0,dots)in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
$endgroup$
add a comment |
$begingroup$
Consider the set $B:={xinell^1:|x|_1<1}$, which is open with respect to $(ell^1,|cdot|)$. We find $mbox{id}B=B$, however $B$ is not open with respect to $(ell^1,|cdot|_s)$, since $0$ is an element, but not an interior point.
Proof: Let $epsilon>0$ and assume without loss of generality that $epsilon<1$. Then $xinell^1$ defined by $x_n=epsilon$ if $nepsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=epsilon$.
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
add a comment |
$begingroup$
Because every continuous open bijection is an homeomorphism.
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $vert xvert_1 le C vert xvert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,dots,1,0,dots)in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
$endgroup$
add a comment |
$begingroup$
If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $vert xvert_1 le C vert xvert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,dots,1,0,dots)in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
$endgroup$
add a comment |
$begingroup$
If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $vert xvert_1 le C vert xvert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,dots,1,0,dots)in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
$endgroup$
If a bijection between topological spaces is open, then its inverse is continuous. In your case this would imply that $vert xvert_1 le C vert xvert_s$ for a uniform constant $C$. Consider the sequence $x^n=(1,dots,1,0,dots)in l^1$ (ones in the first $n$ entries and zeros afterwards) to see that this is wrong.
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
answered Dec 19 '18 at 16:07
Jan BohrJan Bohr
3,3071421
3,3071421
add a comment |
add a comment |
$begingroup$
Consider the set $B:={xinell^1:|x|_1<1}$, which is open with respect to $(ell^1,|cdot|)$. We find $mbox{id}B=B$, however $B$ is not open with respect to $(ell^1,|cdot|_s)$, since $0$ is an element, but not an interior point.
Proof: Let $epsilon>0$ and assume without loss of generality that $epsilon<1$. Then $xinell^1$ defined by $x_n=epsilon$ if $nepsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=epsilon$.
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
add a comment |
$begingroup$
Consider the set $B:={xinell^1:|x|_1<1}$, which is open with respect to $(ell^1,|cdot|)$. We find $mbox{id}B=B$, however $B$ is not open with respect to $(ell^1,|cdot|_s)$, since $0$ is an element, but not an interior point.
Proof: Let $epsilon>0$ and assume without loss of generality that $epsilon<1$. Then $xinell^1$ defined by $x_n=epsilon$ if $nepsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=epsilon$.
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
add a comment |
$begingroup$
Consider the set $B:={xinell^1:|x|_1<1}$, which is open with respect to $(ell^1,|cdot|)$. We find $mbox{id}B=B$, however $B$ is not open with respect to $(ell^1,|cdot|_s)$, since $0$ is an element, but not an interior point.
Proof: Let $epsilon>0$ and assume without loss of generality that $epsilon<1$. Then $xinell^1$ defined by $x_n=epsilon$ if $nepsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=epsilon$.
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
$endgroup$
Consider the set $B:={xinell^1:|x|_1<1}$, which is open with respect to $(ell^1,|cdot|)$. We find $mbox{id}B=B$, however $B$ is not open with respect to $(ell^1,|cdot|_s)$, since $0$ is an element, but not an interior point.
Proof: Let $epsilon>0$ and assume without loss of generality that $epsilon<1$. Then $xinell^1$ defined by $x_n=epsilon$ if $nepsilon<2$ and $x_n=0$ otherwise. Then $|x|_1>1$, however $|x|_s=epsilon$.
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
answered Dec 19 '18 at 16:08
SmileyCraftSmileyCraft
3,591517
3,591517
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
add a comment |
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
Excuse. $B$ is open with respect to $(l^1|.|_1)$ and $B$ is not open with respect to $(l^1,|.|_s)$? I ask it since in the proof, $xin B_ {|. | _ {s}} (0,1)$ and $xnot in B_ {|.| _1} (0,1)$
$endgroup$
– eraldcoil
Dec 19 '18 at 17:24
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
$begingroup$
I'm sorry, but I do not understand what you are trying to say.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:25
add a comment |
$begingroup$
Because every continuous open bijection is an homeomorphism.
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
$endgroup$
add a comment |
$begingroup$
Because every continuous open bijection is an homeomorphism.
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
$endgroup$
add a comment |
$begingroup$
Because every continuous open bijection is an homeomorphism.
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
$endgroup$
Because every continuous open bijection is an homeomorphism.
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
answered Dec 19 '18 at 16:06
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,572917
6,572917
add a comment |
add a comment |
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