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Continuity of a special map between topological sets
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)covering a space with closures of disjoint sets from a basisHow to read a chapter about connectedness for topological spaces as if you only want to know things about metric spaces?Showing that two topologies on the unit circle are the sameUniform continuity of scalar multiplication in topological vector spacesMaps between topological spaces; any rules?$X,Y$ be finite topological spaces such that there exist continuous injections from $X$ to $Y$ and $Y$ to $X$ ; are $X$ and $Y$ homeomorphic?Continuity of inclusion map between subspace and topological spaceHolder continuous map preserves $F_sigma$ setsMetric space vs Uniform space vs Topological SpaceCan the homotopy type change at the limit?
$begingroup$
Let $B_k subset [0,1]^k+1$ and define the map:
$$
phi_k:B_kmapsto C^k[0,1]:(beta_0,beta_1, ldots,beta_k)mapstosum_i=0^kbeta_i b_i,k,
$$
where $b_i,k(t)=binomkit^i(1-t)^k-1,, tin[0,1], i=0,ldots,k$ denotes the Bernstein Polynomial basis of degree $k$. Denote $P_k:=fin C^k[0,1]: f=phi_k(beta_0, ldots,beta_k), (beta_0, ldots,beta_k)in B_k$. Endow $B_k$ and $P_k$ with the Euclidean and the uniform metrics, respectively, denoted by $d_E$ and $d_infty$. Then, it can be readily seen that the map $phi_k:(B_k,d_E)mapsto (P_k, d_infty)$ is continuous.
Now, let me be not very precise for a moment and define a general map $$phi:cup_k=1^infty B_kmapsto cup_k=1^infty P_k$$
such that, if $boldsymbolbetain B_k$, then $phi(boldsymbolbeta)=phi_k(boldsymbolbeta)$. Which type of (metric) topology shoud I use on the "union spaces" so that the map $phi$ is continuous? Should I consider $phi$ as a map between the co-product topological spaces for both $(B_k,d_E)$'s and $(P_k,d_infty)$? Or it is sufficient to consider the co-product topological space for $(B_k,d_E)$'s
and endow $cup_k=1^infty P_k$ with the uniform metric?
real-analysis general-topology functional-analysis metric-spaces parametric
asked Mar 26 at 9:30
Jack LondonJack London
34218
$endgroup$
add a comment |
$begingroup$
Let $B_k subset [0,1]^k+1$ and define the map:
$$
phi_k:B_kmapsto C^k[0,1]:(beta_0,beta_1, ldots,beta_k)mapstosum_i=0^kbeta_i b_i,k,
$$
where $b_i,k(t)=binomkit^i(1-t)^k-1,, tin[0,1], i=0,ldots,k$ denotes the Bernstein Polynomial basis of degree $k$. Denote $P_k:=fin C^k[0,1]: f=phi_k(beta_0, ldots,beta_k), (beta_0, ldots,beta_k)in B_k$. Endow $B_k$ and $P_k$ with the Euclidean and the uniform metrics, respectively, denoted by $d_E$ and $d_infty$. Then, it can be readily seen that the map $phi_k:(B_k,d_E)mapsto (P_k, d_infty)$ is continuous.
Now, let me be not very precise for a moment and define a general map $$phi:cup_k=1^infty B_kmapsto cup_k=1^infty P_k$$
such that, if $boldsymbolbetain B_k$, then $phi(boldsymbolbeta)=phi_k(boldsymbolbeta)$. Which type of (metric) topology shoud I use on the "union spaces" so that the map $phi$ is continuous? Should I consider $phi$ as a map between the co-product topological spaces for both $(B_k,d_E)$'s and $(P_k,d_infty)$? Or it is sufficient to consider the co-product topological space for $(B_k,d_E)$'s
and endow $cup_k=1^infty P_k$ with the uniform metric?
real-analysis general-topology functional-analysis metric-spaces parametric
asked Mar 26 at 9:30
Jack LondonJack London
34218
$endgroup$
$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00
add a comment |
$begingroup$
Let $B_k subset [0,1]^k+1$ and define the map:
$$
phi_k:B_kmapsto C^k[0,1]:(beta_0,beta_1, ldots,beta_k)mapstosum_i=0^kbeta_i b_i,k,
$$
where $b_i,k(t)=binomkit^i(1-t)^k-1,, tin[0,1], i=0,ldots,k$ denotes the Bernstein Polynomial basis of degree $k$. Denote $P_k:=fin C^k[0,1]: f=phi_k(beta_0, ldots,beta_k), (beta_0, ldots,beta_k)in B_k$. Endow $B_k$ and $P_k$ with the Euclidean and the uniform metrics, respectively, denoted by $d_E$ and $d_infty$. Then, it can be readily seen that the map $phi_k:(B_k,d_E)mapsto (P_k, d_infty)$ is continuous.
Now, let me be not very precise for a moment and define a general map $$phi:cup_k=1^infty B_kmapsto cup_k=1^infty P_k$$
such that, if $boldsymbolbetain B_k$, then $phi(boldsymbolbeta)=phi_k(boldsymbolbeta)$. Which type of (metric) topology shoud I use on the "union spaces" so that the map $phi$ is continuous? Should I consider $phi$ as a map between the co-product topological spaces for both $(B_k,d_E)$'s and $(P_k,d_infty)$? Or it is sufficient to consider the co-product topological space for $(B_k,d_E)$'s
and endow $cup_k=1^infty P_k$ with the uniform metric?
real-analysis general-topology functional-analysis metric-spaces parametric
asked Mar 26 at 9:30
Jack LondonJack London
34218
$endgroup$
Let $B_k subset [0,1]^k+1$ and define the map:
$$
phi_k:B_kmapsto C^k[0,1]:(beta_0,beta_1, ldots,beta_k)mapstosum_i=0^kbeta_i b_i,k,
$$
where $b_i,k(t)=binomkit^i(1-t)^k-1,, tin[0,1], i=0,ldots,k$ denotes the Bernstein Polynomial basis of degree $k$. Denote $P_k:=fin C^k[0,1]: f=phi_k(beta_0, ldots,beta_k), (beta_0, ldots,beta_k)in B_k$. Endow $B_k$ and $P_k$ with the Euclidean and the uniform metrics, respectively, denoted by $d_E$ and $d_infty$. Then, it can be readily seen that the map $phi_k:(B_k,d_E)mapsto (P_k, d_infty)$ is continuous.
Now, let me be not very precise for a moment and define a general map $$phi:cup_k=1^infty B_kmapsto cup_k=1^infty P_k$$
such that, if $boldsymbolbetain B_k$, then $phi(boldsymbolbeta)=phi_k(boldsymbolbeta)$. Which type of (metric) topology shoud I use on the "union spaces" so that the map $phi$ is continuous? Should I consider $phi$ as a map between the co-product topological spaces for both $(B_k,d_E)$'s and $(P_k,d_infty)$? Or it is sufficient to consider the co-product topological space for $(B_k,d_E)$'s
and endow $cup_k=1^infty P_k$ with the uniform metric?
real-analysis general-topology functional-analysis metric-spaces parametric
real-analysis general-topology functional-analysis metric-spaces parametric
asked Mar 26 at 9:30
Jack LondonJack London
34218
asked Mar 26 at 9:30
Jack LondonJack London
34218
asked Mar 26 at 9:30
Jack LondonJack London
34218
asked Mar 26 at 9:30
Jack LondonJack London
34218
asked Mar 26 at 9:30
Jack LondonJack London
34218
34218
$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00
add a comment |
$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00
$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00
add a comment |
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$begingroup$
Give $B_k$ the metrics $tilde d(x,y)=fracd(x,y)1+d(x,y)$ where $d$ is the old metric. This doesn't change the topology and bounds your metric by one. Now if you denote the new metric on $B_k$ by $d_k$, defining $$d(x,y)=begincases d_k(x,y) & x,yin B_k\ 2&textelseendcases$$ gives you a metric on $bigcup B_k$ that respects the topology of the components. Note that $P_ksubset C[0,1]$ and thus you can just give the uniform metric to the union of all $P_k$.
$endgroup$
– s.harp
Mar 26 at 9:51
$begingroup$
The tricky thing here is the following. Let $k'>k$. A polynomial $sum_i=0^kbeta_i b_i,k$ can be rewritten as $sum_i=1^k' beta'_ib_i,k'$, for a suitable choice of $beta_0', ldots, beta_k''$. If I correctly get your proposal, we would have $d(boldsymbolbeta,boldsymbolbeta')=2$ and $d_infty(sum_i=0^kbeta_i b_i,k,, sum_i=1^k' beta'_ib_i,k')=0$.
$endgroup$
– Jack London
Mar 26 at 17:23
$begingroup$
Yes, $bigcup P_k$ is not a disjoint union. The map $phi:bigcup B_kto C[0,1]$ is however still continuous. If you want it to be injective then you have to take a quotient of $bigcup B_k$. This may give you trouble with the metric.
$endgroup$
– s.harp
Mar 26 at 17:44
$begingroup$
A possibly stupid question: then I could also look at $d$ as a metric on $overlineB=cup_k=1^infty ktimes B_k$ and $phi$ as a continuous map from $(overlineB,d)$ to $(cup_k=1^infty P_k,d_infty)$, correct?
$endgroup$
– Jack London
Mar 26 at 18:00