Does the rigorous limit definition give rise to the same concept if we consider an $l^p$ metric ($pgeq2$) as...
$begingroup$
Does the $varepsilon!!-!!delta$ limit definition give rise to he same concept if we consider an $l^p$ metric ($pgeq2$) as the distance function?
In other words, do we obtain the same concept and same results if we define limits instead as: for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta implies 0<d_{l^p}(f(x),L)<varepsilon$.
If not, then what about considering instead having the $l^p$ in only one part, like: for every $varepsilon>0$ there is some $delta>0$ such that if $0<|x-a|<deltaimplies 0<d_{l^p}(f(x),L)<varepsilon$. OR for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta$ then $0<|f(x)-L|<varepsilon$.
Furthermore does this apply to the multivariable version of the above definitions?
real-analysis limits
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
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show 1 more comment
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Does the $varepsilon!!-!!delta$ limit definition give rise to he same concept if we consider an $l^p$ metric ($pgeq2$) as the distance function?
In other words, do we obtain the same concept and same results if we define limits instead as: for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta implies 0<d_{l^p}(f(x),L)<varepsilon$.
If not, then what about considering instead having the $l^p$ in only one part, like: for every $varepsilon>0$ there is some $delta>0$ such that if $0<|x-a|<deltaimplies 0<d_{l^p}(f(x),L)<varepsilon$. OR for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta$ then $0<|f(x)-L|<varepsilon$.
Furthermore does this apply to the multivariable version of the above definitions?
real-analysis limits
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
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@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
1
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
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@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01
|
show 1 more comment
$begingroup$
Does the $varepsilon!!-!!delta$ limit definition give rise to he same concept if we consider an $l^p$ metric ($pgeq2$) as the distance function?
In other words, do we obtain the same concept and same results if we define limits instead as: for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta implies 0<d_{l^p}(f(x),L)<varepsilon$.
If not, then what about considering instead having the $l^p$ in only one part, like: for every $varepsilon>0$ there is some $delta>0$ such that if $0<|x-a|<deltaimplies 0<d_{l^p}(f(x),L)<varepsilon$. OR for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta$ then $0<|f(x)-L|<varepsilon$.
Furthermore does this apply to the multivariable version of the above definitions?
real-analysis limits
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
Does the $varepsilon!!-!!delta$ limit definition give rise to he same concept if we consider an $l^p$ metric ($pgeq2$) as the distance function?
In other words, do we obtain the same concept and same results if we define limits instead as: for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta implies 0<d_{l^p}(f(x),L)<varepsilon$.
If not, then what about considering instead having the $l^p$ in only one part, like: for every $varepsilon>0$ there is some $delta>0$ such that if $0<|x-a|<deltaimplies 0<d_{l^p}(f(x),L)<varepsilon$. OR for every $varepsilon>0$ there is some $delta>0$ such that if $0<d_{l^p}(x,a)<delta$ then $0<|f(x)-L|<varepsilon$.
Furthermore does this apply to the multivariable version of the above definitions?
real-analysis limits
real-analysis limits
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
edited Jan 31 at 7:59
Stupid Questions Inc
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
asked Jan 20 at 14:25
Stupid Questions IncStupid Questions Inc
7010
7010
$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
$begingroup$
@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
1
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
$begingroup$
@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01
|
show 1 more comment
$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
$begingroup$
@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
1
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
$begingroup$
@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01
$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
$begingroup$
@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
$begingroup$
@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
1
1
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
$begingroup$
@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01
|
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$begingroup$
What do you men by concept ? Do you mean it induces the same topology ?
$endgroup$
– J.F
Jan 20 at 14:30
$begingroup$
@G.F No, I mean does it yield the same limit concept, i.e. will we get the same results if we use the new limit formulation that I mentioned?
$endgroup$
– Stupid Questions Inc
Jan 21 at 12:41
1
$begingroup$
In general, different metrics (topologies) yield different limits unless they are equivalent metrics (topologies). What space are you on btw? $mathbb{R}$? (Usually $l^p$ denotes the metric on the space of sequences, but maybe you meant $L^p$)
$endgroup$
– Harnak
Jan 22 at 8:56
$begingroup$
@Harnak Yes $mathbb{R}$ where $l^p$ is $(|x_i-y_i|^p+cdots+|x_n-y_n|^p)^{1/p}$. If that's the case then I'd be interested in some example, also what about the second part of my question where only one part of the implication is formulated using the $l^p$ distance?
$endgroup$
– Stupid Questions Inc
Jan 25 at 8:29
$begingroup$
This is true. Also for multivariable, i.e. $mathbb{R}^n$. The topologies induced by the $p$- norm, and euclidean norm are equivalent. You can show that all the open sets in one topology are open in the other topology. It's enough to show that every open ball in one topology contains an open ball in the other topology, I'll leave this to you.
$endgroup$
– Kolja
Jan 31 at 18:01