How to show that $g:2^Mto 2^mathbb{N}$ defined by $g(A) = Xcup A$ is continuous?
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In Galvin and Prikry's paper, they inroduce completely Ramsey sets.
Definition $5$: A set $Ssubseteq 2^mathbb{N}$ is completely Ramsey if $f^{-1}(S)$ is Ramsey for every continuous mapping $f:2^mathbb{N}to 2^mathbb{N}.$
Question: What is a topology in $2^mathbb{N}?$
As mentioned by @Patrick Stevens below, the authors consider product topology on $2^mathbb{N}.$
I have another question:
Question: Suppose that $X$ is a finite subset of $mathbb{N}$ and $M$ is a countable subset of $mathbb{N}.$
Define $g:2^Mto 2^mathbb{N}$ by
$$g(A) = Xcup A.$$
How to show that $g$ is continuous?
The function $g$ is defined in the proof of Lemma $7$ of the paper above.
The authors do not prove it.
general-topology descriptive-set-theory ramsey-theory
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In Galvin and Prikry's paper, they inroduce completely Ramsey sets.
Definition $5$: A set $Ssubseteq 2^mathbb{N}$ is completely Ramsey if $f^{-1}(S)$ is Ramsey for every continuous mapping $f:2^mathbb{N}to 2^mathbb{N}.$
Question: What is a topology in $2^mathbb{N}?$
As mentioned by @Patrick Stevens below, the authors consider product topology on $2^mathbb{N}.$
I have another question:
Question: Suppose that $X$ is a finite subset of $mathbb{N}$ and $M$ is a countable subset of $mathbb{N}.$
Define $g:2^Mto 2^mathbb{N}$ by
$$g(A) = Xcup A.$$
How to show that $g$ is continuous?
The function $g$ is defined in the proof of Lemma $7$ of the paper above.
The authors do not prove it.
general-topology descriptive-set-theory ramsey-theory
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In Galvin and Prikry's paper, they inroduce completely Ramsey sets.
Definition $5$: A set $Ssubseteq 2^mathbb{N}$ is completely Ramsey if $f^{-1}(S)$ is Ramsey for every continuous mapping $f:2^mathbb{N}to 2^mathbb{N}.$
Question: What is a topology in $2^mathbb{N}?$
As mentioned by @Patrick Stevens below, the authors consider product topology on $2^mathbb{N}.$
I have another question:
Question: Suppose that $X$ is a finite subset of $mathbb{N}$ and $M$ is a countable subset of $mathbb{N}.$
Define $g:2^Mto 2^mathbb{N}$ by
$$g(A) = Xcup A.$$
How to show that $g$ is continuous?
The function $g$ is defined in the proof of Lemma $7$ of the paper above.
The authors do not prove it.
general-topology descriptive-set-theory ramsey-theory
In Galvin and Prikry's paper, they inroduce completely Ramsey sets.
Definition $5$: A set $Ssubseteq 2^mathbb{N}$ is completely Ramsey if $f^{-1}(S)$ is Ramsey for every continuous mapping $f:2^mathbb{N}to 2^mathbb{N}.$
Question: What is a topology in $2^mathbb{N}?$
As mentioned by @Patrick Stevens below, the authors consider product topology on $2^mathbb{N}.$
I have another question:
Question: Suppose that $X$ is a finite subset of $mathbb{N}$ and $M$ is a countable subset of $mathbb{N}.$
Define $g:2^Mto 2^mathbb{N}$ by
$$g(A) = Xcup A.$$
How to show that $g$ is continuous?
The function $g$ is defined in the proof of Lemma $7$ of the paper above.
The authors do not prove it.
general-topology descriptive-set-theory ramsey-theory
general-topology descriptive-set-theory ramsey-theory
edited Nov 17 at 11:23
Henno Brandsma
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102k344108
asked Nov 17 at 9:36
Idonknow
2,254746111
2,254746111
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2 Answers
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A map into a product is continuous iff all the compositions with the projections from that product are continuous.
Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:
Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.
Finiteness of $X$ nor infiniteness of $M$ plays any part.
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
add a comment |
up vote
1
down vote
As the paper says in the first paragraph, it's the product topology.
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A map into a product is continuous iff all the compositions with the projections from that product are continuous.
Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:
Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.
Finiteness of $X$ nor infiniteness of $M$ plays any part.
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
add a comment |
up vote
2
down vote
accepted
A map into a product is continuous iff all the compositions with the projections from that product are continuous.
Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:
Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.
Finiteness of $X$ nor infiniteness of $M$ plays any part.
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A map into a product is continuous iff all the compositions with the projections from that product are continuous.
Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:
Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.
Finiteness of $X$ nor infiniteness of $M$ plays any part.
A map into a product is continuous iff all the compositions with the projections from that product are continuous.
Now $pi_n circ g$ is the constant $1$ map for $n in X$, the identity for all other $n$, so always continuous, so $g$ is too:
Note that $A$ is just identified with $chi_A: mathbb{N} to 2$, so $n in A$ is just the same as $pi_n(A)=1$. So $pi_n(A cup X)$ is $1$ for all $n in X$ and all $n in A$ and $0$ for all $ n notin A cup X$. So when $n notin X$, $pi_n(g(A))= pi_n(A cup X) = pi_n(A)$ and for all $n in X$, $pi_n(X cup A) = 1$.
Finiteness of $X$ nor infiniteness of $M$ plays any part.
edited Nov 17 at 10:03
answered Nov 17 at 9:51
Henno Brandsma
102k344108
102k344108
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
add a comment |
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
It seems that from your argument above, any function $g:2^Mto 2^mathbb{N}$ is continuous. May I know where is my mistake? In particular, I do not see the definition of $g$ being used anywhere in your argument.
– Idonknow
Nov 17 at 9:56
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
@Idonknow No, I use the specific definition of $g$ to see the identities on the projections. There are many discontinuous maps from $2^M$ to $2^mathbb{N}$.
– Henno Brandsma
Nov 17 at 9:58
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
How do you obtain that $pi_ncirc g$ is the constant function $1$?
– Idonknow
Nov 17 at 10:00
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
@Idonknow I expanded the argument. $X$ is always a subset of $g(A)$.
– Henno Brandsma
Nov 17 at 10:04
add a comment |
up vote
1
down vote
As the paper says in the first paragraph, it's the product topology.
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
add a comment |
up vote
1
down vote
As the paper says in the first paragraph, it's the product topology.
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
add a comment |
up vote
1
down vote
up vote
1
down vote
As the paper says in the first paragraph, it's the product topology.
As the paper says in the first paragraph, it's the product topology.
answered Nov 17 at 9:37
Patrick Stevens
28.1k52874
28.1k52874
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
add a comment |
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
I have modified my question. In particular, how do we show that a map $g:2^Mto 2^mathbb{N}$ is continuous where $M$ is a countable subset of $mathbb{N}?$
– Idonknow
Nov 17 at 9:47
add a comment |
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