Step and piecewise continuous linear function on $[0,1]$ are separable? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Prove integral metric is separableAbout density of some piecewise linear function in some subspace of $C[a,b]$If $f_ncolon [0, 1] to [0, 1]$ are nondecreasing and $f_n$ converges pointwise to a continuous $f$, then the convergence is uniformSeparability of a product metric spacePiecewise linear function close to continuous functionThe Baire space $mathscrN$ is separableStep functions on $[a,b]$ are dense in $mathcal C^0([a,b])$.Show $f(x)=sqrtx$ is continuous on $[0,1]$How can I prove $d_1(x,y) leq n d_infty (x,y)$A normed space is not separable when $ell_infty$ can be embedded isometrically into itContinuous function which is not piecewise monotonous
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Step and piecewise continuous linear function on $[0,1]$ are separable?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Prove integral metric is separableAbout density of some piecewise linear function in some subspace of $C[a,b]$If $f_ncolon [0, 1] to [0, 1]$ are nondecreasing and $f_n$ converges pointwise to a continuous $f$, then the convergence is uniformSeparability of a product metric spacePiecewise linear function close to continuous functionThe Baire space $mathscrN$ is separableStep functions on $[a,b]$ are dense in $mathcal C^0([a,b])$.Show $f(x)=sqrtx$ is continuous on $[0,1]$How can I prove $d_1(x,y) leq n d_infty (x,y)$A normed space is not separable when $ell_infty$ can be embedded isometrically into itContinuous function which is not piecewise monotonous
$begingroup$
Given $E[0,1]$ be the set of all step functions on $[0,1]$ and $L[0,1]$ be the set of all piecewise linear continuous functions on $[0,1]$.
Then
(a) $(E [0,1], d_infty)$ is separable?
(b) $(E [0,1], d_1)$ is separable?
(c) $(L [0,1], d_infty)$ is separable?
First all, I am using the following definitions:
1) $(X,d)$ is called a separable metric space if it contains a countable, dense subset.
2) $d_infty: X times Xrightarrow mathbbR$ given by $d_infty(f,g) = sup_x in [0,1]|f(x)-g(x)|$.
3) $d_1 : X times Xrightarrow mathbbR$ given by $d_1(f,g) = int_0^1|f(x)-g(x)|;dx$.
4) $E[0,1]$ is the set of all functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f$ is constant in all open subinterval $(x_i, x_i+1), i = 0, ldots, n$.
5) $L[0,1]$ is the set of all continuous functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f(x) = f(x_i)+
dfrac(f(x_i+1) - f(x_i))x_i+1-x_i(x-x_i)$ if $x in [x_i, x_i+1]$, $0 leq i leq n$.
edit Problem solved.
real-analysis analysis metric-spaces
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
$endgroup$
|
show 1 more comment
$begingroup$
Given $E[0,1]$ be the set of all step functions on $[0,1]$ and $L[0,1]$ be the set of all piecewise linear continuous functions on $[0,1]$.
Then
(a) $(E [0,1], d_infty)$ is separable?
(b) $(E [0,1], d_1)$ is separable?
(c) $(L [0,1], d_infty)$ is separable?
First all, I am using the following definitions:
1) $(X,d)$ is called a separable metric space if it contains a countable, dense subset.
2) $d_infty: X times Xrightarrow mathbbR$ given by $d_infty(f,g) = sup_x in [0,1]|f(x)-g(x)|$.
3) $d_1 : X times Xrightarrow mathbbR$ given by $d_1(f,g) = int_0^1|f(x)-g(x)|;dx$.
4) $E[0,1]$ is the set of all functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f$ is constant in all open subinterval $(x_i, x_i+1), i = 0, ldots, n$.
5) $L[0,1]$ is the set of all continuous functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f(x) = f(x_i)+
dfrac(f(x_i+1) - f(x_i))x_i+1-x_i(x-x_i)$ if $x in [x_i, x_i+1]$, $0 leq i leq n$.
edit Problem solved.
real-analysis analysis metric-spaces
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
$endgroup$
$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30
|
show 1 more comment
$begingroup$
Given $E[0,1]$ be the set of all step functions on $[0,1]$ and $L[0,1]$ be the set of all piecewise linear continuous functions on $[0,1]$.
Then
(a) $(E [0,1], d_infty)$ is separable?
(b) $(E [0,1], d_1)$ is separable?
(c) $(L [0,1], d_infty)$ is separable?
First all, I am using the following definitions:
1) $(X,d)$ is called a separable metric space if it contains a countable, dense subset.
2) $d_infty: X times Xrightarrow mathbbR$ given by $d_infty(f,g) = sup_x in [0,1]|f(x)-g(x)|$.
3) $d_1 : X times Xrightarrow mathbbR$ given by $d_1(f,g) = int_0^1|f(x)-g(x)|;dx$.
4) $E[0,1]$ is the set of all functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f$ is constant in all open subinterval $(x_i, x_i+1), i = 0, ldots, n$.
5) $L[0,1]$ is the set of all continuous functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f(x) = f(x_i)+
dfrac(f(x_i+1) - f(x_i))x_i+1-x_i(x-x_i)$ if $x in [x_i, x_i+1]$, $0 leq i leq n$.
edit Problem solved.
real-analysis analysis metric-spaces
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
$endgroup$
Given $E[0,1]$ be the set of all step functions on $[0,1]$ and $L[0,1]$ be the set of all piecewise linear continuous functions on $[0,1]$.
Then
(a) $(E [0,1], d_infty)$ is separable?
(b) $(E [0,1], d_1)$ is separable?
(c) $(L [0,1], d_infty)$ is separable?
First all, I am using the following definitions:
1) $(X,d)$ is called a separable metric space if it contains a countable, dense subset.
2) $d_infty: X times Xrightarrow mathbbR$ given by $d_infty(f,g) = sup_x in [0,1]|f(x)-g(x)|$.
3) $d_1 : X times Xrightarrow mathbbR$ given by $d_1(f,g) = int_0^1|f(x)-g(x)|;dx$.
4) $E[0,1]$ is the set of all functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f$ is constant in all open subinterval $(x_i, x_i+1), i = 0, ldots, n$.
5) $L[0,1]$ is the set of all continuous functions $f:[0,1] rightarrow mathbbR$ such that there are $0 = x_0 < x_1 < cdots < x_n < x_n+1=1$, $n geq 0$, where $f(x) = f(x_i)+
dfrac(f(x_i+1) - f(x_i))x_i+1-x_i(x-x_i)$ if $x in [x_i, x_i+1]$, $0 leq i leq n$.
edit Problem solved.
real-analysis analysis metric-spaces
real-analysis analysis metric-spaces
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
edited Apr 3 at 14:54
Thiago Alexandre
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
asked Apr 1 at 19:13
Thiago AlexandreThiago Alexandre
1347
1347
$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30
|
show 1 more comment
$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30
$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $$mathscr P_n=0=x_0<x_1<cdots<x_n<x_n+1=1:x_iinmathbb Q,$$ let $$mathscr P=bigcup_n=0^inftymathscr P_n,$$ let $$mathscr Q_P=(x_i,x_i+1)_i=0^ntext for every P=0=x_0<x_1<cdots<x_n<x_n+1=1inmathscr P,$$ and let $$S=f:[0,1]tomathbb Q:Pinmathscr Ptext and ftext is constant on every Iinmathscr Q_P.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]supset E[0,1]$ is separable and thus $E[0,1]$ is separable.
answered Apr 2 at 0:18
JosuéJosué
3,49742672
$endgroup$
1
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
add a comment |
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1 Answer
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$begingroup$
Let $$mathscr P_n=0=x_0<x_1<cdots<x_n<x_n+1=1:x_iinmathbb Q,$$ let $$mathscr P=bigcup_n=0^inftymathscr P_n,$$ let $$mathscr Q_P=(x_i,x_i+1)_i=0^ntext for every P=0=x_0<x_1<cdots<x_n<x_n+1=1inmathscr P,$$ and let $$S=f:[0,1]tomathbb Q:Pinmathscr Ptext and ftext is constant on every Iinmathscr Q_P.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]supset E[0,1]$ is separable and thus $E[0,1]$ is separable.
answered Apr 2 at 0:18
JosuéJosué
3,49742672
$endgroup$
1
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
add a comment |
$begingroup$
Let $$mathscr P_n=0=x_0<x_1<cdots<x_n<x_n+1=1:x_iinmathbb Q,$$ let $$mathscr P=bigcup_n=0^inftymathscr P_n,$$ let $$mathscr Q_P=(x_i,x_i+1)_i=0^ntext for every P=0=x_0<x_1<cdots<x_n<x_n+1=1inmathscr P,$$ and let $$S=f:[0,1]tomathbb Q:Pinmathscr Ptext and ftext is constant on every Iinmathscr Q_P.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]supset E[0,1]$ is separable and thus $E[0,1]$ is separable.
answered Apr 2 at 0:18
JosuéJosué
3,49742672
$endgroup$
1
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
add a comment |
$begingroup$
Let $$mathscr P_n=0=x_0<x_1<cdots<x_n<x_n+1=1:x_iinmathbb Q,$$ let $$mathscr P=bigcup_n=0^inftymathscr P_n,$$ let $$mathscr Q_P=(x_i,x_i+1)_i=0^ntext for every P=0=x_0<x_1<cdots<x_n<x_n+1=1inmathscr P,$$ and let $$S=f:[0,1]tomathbb Q:Pinmathscr Ptext and ftext is constant on every Iinmathscr Q_P.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]supset E[0,1]$ is separable and thus $E[0,1]$ is separable.
answered Apr 2 at 0:18
JosuéJosué
3,49742672
$endgroup$
Let $$mathscr P_n=0=x_0<x_1<cdots<x_n<x_n+1=1:x_iinmathbb Q,$$ let $$mathscr P=bigcup_n=0^inftymathscr P_n,$$ let $$mathscr Q_P=(x_i,x_i+1)_i=0^ntext for every P=0=x_0<x_1<cdots<x_n<x_n+1=1inmathscr P,$$ and let $$S=f:[0,1]tomathbb Q:Pinmathscr Ptext and ftext is constant on every Iinmathscr Q_P.$$ Can you show that $S$ is a countable, dense subset of $E[0,1]$ with respect to the $d_1$ metric?
Alternatively, you can show that $[0,1]$ is a separable measure space, so that $L^1[0,1]supset E[0,1]$ is separable and thus $E[0,1]$ is separable.
answered Apr 2 at 0:18
JosuéJosué
3,49742672
edited Apr 2 at 0:36
answered Apr 2 at 0:18
JosuéJosué
3,49742672
answered Apr 2 at 0:18
JosuéJosué
3,49742672
answered Apr 2 at 0:18
JosuéJosué
3,49742672
3,49742672
1
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$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
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– Thiago Alexandre
Apr 2 at 0:50
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
add a comment |
1
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
1
1
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
$begingroup$
$S$ is countable because it is taken on all countable partitions $P$ and $S$ is dense because given $epsilon > 0$, $g in E[0,1]$, there is a $f in S$ such that $int_0^1|f(x)-g(x)|;dx < epsilon$. This is because $g$ has a finite partition $P$ and there is a partition $P_n$ of rational ones that approximate $f$ of $g$ making the integral as small as it wants. Is it make sense?
$endgroup$
– Thiago Alexandre
Apr 2 at 0:50
1
1
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
$begingroup$
I got it. For any step function $f in E[0,1]$ there is a partition $P$ that approaches by rational numbers for the partition of $f$ function. So, we have $g in S$ such that $d_1(f,g) = |c_1 - q_1| + |c_2 - q_2| + cdots + |c_n+1-q_n+1| < epsilon$ because each parcel |c_i - q_i| we can get that is smaller than $fracepsilonn+1$.
$endgroup$
– Thiago Alexandre
Apr 3 at 14:42
add a comment |
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$begingroup$
See tinyurl.com/y2o2t3ky for why $(E([0,1]),d_infty)$ is not separable.
$endgroup$
– Josué
Apr 1 at 20:58
$begingroup$
@Cleric I got it. We take $Y subset E[0,1]$ with $d_infty (f_t, f_s) = 1, forall t neq s$. How Y is uncountable, then $(E[0,1], d_infty)$ is not separable.
$endgroup$
– Thiago Alexandre
Apr 1 at 21:58
$begingroup$
Your argument for (b) is wrong. I think you misled yourself by writing $|t-s|=r$, because the distance between $f_t$ and $f_s$ varies with $t$ and $s$, and in particular can become arbitrarily small. To show a set is not separable by this argument, you must find a fixed constant $c$ and an uncountable set of elements for which any pair is more than distance $c$ apart. You have not done that, and in fact it is not possible; $E[0,1]$ is actually separable with respect to the $d_1$ metric.
$endgroup$
– Nate Eldredge
Apr 1 at 23:15
$begingroup$
@NateEldredge Thanks. I understand my wrong. So I need to find a subset dense and countable in $(E[0,1], d_1)$.
$endgroup$
– Thiago Alexandre
Apr 1 at 23:27
$begingroup$
@NateEldredge I found this result math.stackexchange.com/questions/195070/…
$endgroup$
– Thiago Alexandre
Apr 1 at 23:30