Show that $d_infty: l^infty times l^infty longrightarrow mathbbR^+$ is well defined. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the set $E$ of sequences containing only entries $0$ and $1$ in $(m,left | cdot right |_infty)$ complete?Definition of a metricHow to prove a map between two spaces of real sequences $f : l^1 to l^2 $ is well-defined and continuousWhat does it REALLY mean for a metric space to be compact?Prove that the given set is a metric space?Show that $d: X times X rightarrow mathbbR$ is a metric on $X$A function f on a metric space is uniformly continuous iff for every pair of subsets A, B with d(A, B)=0 implies that d'(f(A), f(B))=0Definition of CompactnessShow that $f:Xto Y$ is uniformly continuous $iff f(N_delta,X(x))subset N_epsilon,Y(f(x))$Problem with proof of every open subset of the real line is a countable union of disjoint open intervals.
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Show that $d_infty: l^infty times l^infty longrightarrow mathbbR^+$ is well defined.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the set $E$ of sequences containing only entries $0$ and $1$ in $(m,left | cdot right |_infty)$ complete?Definition of a metricHow to prove a map between two spaces of real sequences $f : l^1 to l^2 $ is well-defined and continuousWhat does it REALLY mean for a metric space to be compact?Prove that the given set is a metric space?Show that $d: X times X rightarrow mathbbR$ is a metric on $X$A function f on a metric space is uniformly continuous iff for every pair of subsets A, B with d(A, B)=0 implies that d'(f(A), f(B))=0Definition of CompactnessShow that $f:Xto Y$ is uniformly continuous $iff f(N_delta,X(x))subset N_epsilon,Y(f(x))$Problem with proof of every open subset of the real line is a countable union of disjoint open intervals.
$begingroup$
I have come across some answers as to how to prove a function is well-defined. Problem is, I can't wrap my head around the fact that the definition itself here implies that the function is well-defined.
By proving this is a metric on $l^infty$ wouldn't I have proved it's well defined anyway?
analysis metric-spaces
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
$endgroup$
add a comment |
$begingroup$
I have come across some answers as to how to prove a function is well-defined. Problem is, I can't wrap my head around the fact that the definition itself here implies that the function is well-defined.
By proving this is a metric on $l^infty$ wouldn't I have proved it's well defined anyway?
analysis metric-spaces
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
$endgroup$
1
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01
add a comment |
$begingroup$
I have come across some answers as to how to prove a function is well-defined. Problem is, I can't wrap my head around the fact that the definition itself here implies that the function is well-defined.
By proving this is a metric on $l^infty$ wouldn't I have proved it's well defined anyway?
analysis metric-spaces
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
$endgroup$
I have come across some answers as to how to prove a function is well-defined. Problem is, I can't wrap my head around the fact that the definition itself here implies that the function is well-defined.
By proving this is a metric on $l^infty$ wouldn't I have proved it's well defined anyway?
analysis metric-spaces
analysis metric-spaces
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
edited Mar 25 at 19:36
upStoneLock
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
asked Mar 25 at 19:33
upStoneLockupStoneLock
204
204
1
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01
add a comment |
1
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01
1
1
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01
add a comment |
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1
$begingroup$
Well defined means it indeed does the job it's built for and an argument belonging in $ell^infty times ell^infty$ indeed ends up as a value in $mathbb R^+$.
$endgroup$
– Rebellos
Mar 25 at 19:41
$begingroup$
@Rebellos thanks! That does clarify things a bit!
$endgroup$
– upStoneLock
Mar 25 at 21:01