Mixing
$C(X)$ is infinite-dimensional, where $X$ is infinite compact Hausdorff space
Clash Royale CLAN TAG#URR8PPP
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Could you tell me where is mistake in my proof?
Suppose that exist finite basis $e_1,dots ,e_N$. Then $q(x)=a_1e_1+dots +a_Ne_N$ where $q(x) = q$ for every $x$, where $q$ is number. Take some $f$ from $C(X)$ then $$ f(x_0)=r=a_1e_1(x_0)+dots +a_Ne_N(x_0)=a_1e_1+dots +a_Ne_N=f(x)$$ for every $x in X$. That implies that $ f $ is a constant function so every function in $C(X)$ is constant. But by Urysohn's lemma I can find a continous function s.t $f(x)=1$ and $f(y)=0$.
functional-analysis proof-verification
edited Jul 22 at 21:05
mechanodroid
22.2k52041
asked Jul 22 at 18:00
user577360
331
add a comment |Â
up vote
2
down vote
favorite
Could you tell me where is mistake in my proof?
Suppose that exist finite basis $e_1,dots ,e_N$. Then $q(x)=a_1e_1+dots +a_Ne_N$ where $q(x) = q$ for every $x$, where $q$ is number. Take some $f$ from $C(X)$ then $$ f(x_0)=r=a_1e_1(x_0)+dots +a_Ne_N(x_0)=a_1e_1+dots +a_Ne_N=f(x)$$ for every $x in X$. That implies that $ f $ is a constant function so every function in $C(X)$ is constant. But by Urysohn's lemma I can find a continous function s.t $f(x)=1$ and $f(y)=0$.
functional-analysis proof-verification
edited Jul 22 at 21:05
mechanodroid
22.2k52041
asked Jul 22 at 18:00
user577360
331
4
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
1
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Could you tell me where is mistake in my proof?
Suppose that exist finite basis $e_1,dots ,e_N$. Then $q(x)=a_1e_1+dots +a_Ne_N$ where $q(x) = q$ for every $x$, where $q$ is number. Take some $f$ from $C(X)$ then $$ f(x_0)=r=a_1e_1(x_0)+dots +a_Ne_N(x_0)=a_1e_1+dots +a_Ne_N=f(x)$$ for every $x in X$. That implies that $ f $ is a constant function so every function in $C(X)$ is constant. But by Urysohn's lemma I can find a continous function s.t $f(x)=1$ and $f(y)=0$.
functional-analysis proof-verification
edited Jul 22 at 21:05
mechanodroid
22.2k52041
asked Jul 22 at 18:00
user577360
331
Could you tell me where is mistake in my proof?
Suppose that exist finite basis $e_1,dots ,e_N$. Then $q(x)=a_1e_1+dots +a_Ne_N$ where $q(x) = q$ for every $x$, where $q$ is number. Take some $f$ from $C(X)$ then $$ f(x_0)=r=a_1e_1(x_0)+dots +a_Ne_N(x_0)=a_1e_1+dots +a_Ne_N=f(x)$$ for every $x in X$. That implies that $ f $ is a constant function so every function in $C(X)$ is constant. But by Urysohn's lemma I can find a continous function s.t $f(x)=1$ and $f(y)=0$.
functional-analysis proof-verification
edited Jul 22 at 21:05
mechanodroid
22.2k52041
asked Jul 22 at 18:00
user577360
331
edited Jul 22 at 21:05
mechanodroid
22.2k52041
edited Jul 22 at 21:05
mechanodroid
22.2k52041
edited Jul 22 at 21:05
mechanodroid
22.2k52041
22.2k52041
asked Jul 22 at 18:00
user577360
331
asked Jul 22 at 18:00
user577360
331
asked Jul 22 at 18:00
user577360
331
331
4
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
1
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16
add a comment |Â
4
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
1
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16
4
4
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
1
1
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16
add a comment |Â
1 Answer
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0
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The mistake is that you seem to assume that $e_i$ are constant functions, which doesn't have to be true.
You can try this.
Inductively construct a sequence of disjoint nonempty open sets $(V_n)_n$ like this:
Take $x, y in X$, $x ne y$. There exist disjoint open sets $U, V$ such that $x in U$, $y in V$. In particular $Xsetminus overlineU ne emptyset$.
If $Xsetminus overlineU$ is infinite, set $V_1 = U$.
On the other hand, if $Xsetminus overlineU$ is finite, set $V_1 = Xsetminus overlineU$. Then $V_1$ is a finite nonempty open set so it is also closed and hence $Xsetminus overlineV_1 = X setminus V_1$ is infinite.
In either case, $Xsetminus overlineV_1$ is infinite so we can inductively continue with the compact Hausdorff space $Xsetminus overlineV_1$.
Now pick $x_n in V_n$. By Urysohn's lemma there $exists f_n in C(X)$ such that $f(x_n) = 1$ and $operatornamesupp f_n subseteq V_n$.
Since $f_n_ninmathbbN$ are nonzero and have disjoint supports, they are clearly linearly independent so $C(X)$ is infinite-dimensional.
answered Jul 22 at 21:03
mechanodroid
22.2k52041
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The mistake is that you seem to assume that $e_i$ are constant functions, which doesn't have to be true.
You can try this.
Inductively construct a sequence of disjoint nonempty open sets $(V_n)_n$ like this:
Take $x, y in X$, $x ne y$. There exist disjoint open sets $U, V$ such that $x in U$, $y in V$. In particular $Xsetminus overlineU ne emptyset$.
If $Xsetminus overlineU$ is infinite, set $V_1 = U$.
On the other hand, if $Xsetminus overlineU$ is finite, set $V_1 = Xsetminus overlineU$. Then $V_1$ is a finite nonempty open set so it is also closed and hence $Xsetminus overlineV_1 = X setminus V_1$ is infinite.
In either case, $Xsetminus overlineV_1$ is infinite so we can inductively continue with the compact Hausdorff space $Xsetminus overlineV_1$.
Now pick $x_n in V_n$. By Urysohn's lemma there $exists f_n in C(X)$ such that $f(x_n) = 1$ and $operatornamesupp f_n subseteq V_n$.
Since $f_n_ninmathbbN$ are nonzero and have disjoint supports, they are clearly linearly independent so $C(X)$ is infinite-dimensional.
answered Jul 22 at 21:03
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
The mistake is that you seem to assume that $e_i$ are constant functions, which doesn't have to be true.
You can try this.
Inductively construct a sequence of disjoint nonempty open sets $(V_n)_n$ like this:
Take $x, y in X$, $x ne y$. There exist disjoint open sets $U, V$ such that $x in U$, $y in V$. In particular $Xsetminus overlineU ne emptyset$.
If $Xsetminus overlineU$ is infinite, set $V_1 = U$.
On the other hand, if $Xsetminus overlineU$ is finite, set $V_1 = Xsetminus overlineU$. Then $V_1$ is a finite nonempty open set so it is also closed and hence $Xsetminus overlineV_1 = X setminus V_1$ is infinite.
In either case, $Xsetminus overlineV_1$ is infinite so we can inductively continue with the compact Hausdorff space $Xsetminus overlineV_1$.
Now pick $x_n in V_n$. By Urysohn's lemma there $exists f_n in C(X)$ such that $f(x_n) = 1$ and $operatornamesupp f_n subseteq V_n$.
Since $f_n_ninmathbbN$ are nonzero and have disjoint supports, they are clearly linearly independent so $C(X)$ is infinite-dimensional.
answered Jul 22 at 21:03
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The mistake is that you seem to assume that $e_i$ are constant functions, which doesn't have to be true.
You can try this.
Inductively construct a sequence of disjoint nonempty open sets $(V_n)_n$ like this:
Take $x, y in X$, $x ne y$. There exist disjoint open sets $U, V$ such that $x in U$, $y in V$. In particular $Xsetminus overlineU ne emptyset$.
If $Xsetminus overlineU$ is infinite, set $V_1 = U$.
On the other hand, if $Xsetminus overlineU$ is finite, set $V_1 = Xsetminus overlineU$. Then $V_1$ is a finite nonempty open set so it is also closed and hence $Xsetminus overlineV_1 = X setminus V_1$ is infinite.
In either case, $Xsetminus overlineV_1$ is infinite so we can inductively continue with the compact Hausdorff space $Xsetminus overlineV_1$.
Now pick $x_n in V_n$. By Urysohn's lemma there $exists f_n in C(X)$ such that $f(x_n) = 1$ and $operatornamesupp f_n subseteq V_n$.
Since $f_n_ninmathbbN$ are nonzero and have disjoint supports, they are clearly linearly independent so $C(X)$ is infinite-dimensional.
answered Jul 22 at 21:03
mechanodroid
22.2k52041
The mistake is that you seem to assume that $e_i$ are constant functions, which doesn't have to be true.
You can try this.
Inductively construct a sequence of disjoint nonempty open sets $(V_n)_n$ like this:
Take $x, y in X$, $x ne y$. There exist disjoint open sets $U, V$ such that $x in U$, $y in V$. In particular $Xsetminus overlineU ne emptyset$.
If $Xsetminus overlineU$ is infinite, set $V_1 = U$.
On the other hand, if $Xsetminus overlineU$ is finite, set $V_1 = Xsetminus overlineU$. Then $V_1$ is a finite nonempty open set so it is also closed and hence $Xsetminus overlineV_1 = X setminus V_1$ is infinite.
In either case, $Xsetminus overlineV_1$ is infinite so we can inductively continue with the compact Hausdorff space $Xsetminus overlineV_1$.
Now pick $x_n in V_n$. By Urysohn's lemma there $exists f_n in C(X)$ such that $f(x_n) = 1$ and $operatornamesupp f_n subseteq V_n$.
Since $f_n_ninmathbbN$ are nonzero and have disjoint supports, they are clearly linearly independent so $C(X)$ is infinite-dimensional.
answered Jul 22 at 21:03
mechanodroid
22.2k52041
answered Jul 22 at 21:03
mechanodroid
22.2k52041
answered Jul 22 at 21:03
mechanodroid
22.2k52041
answered Jul 22 at 21:03
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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4
Why is this getting downvoted? It's a proof verification. There's clear effort from the OP.
– Shaun
Jul 22 at 18:07
I didn't downvote, but I can hardly understand the proof.
– Kenny Lau
Jul 22 at 18:15
1
It seems to me that OP is assuming $e_i$ are numbers instead of functions.
– Kenny Lau
Jul 22 at 18:16