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Given continuous $f:0,1^Sto[0,1]$ with S uncountable and $f(sigma)=1$, show that $f(tau)=1$ for some $tauneqsigma$
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Let $S$ be an uncountable set and consider the space $0,1^S$ under the product topology where $0,1$ is discrete. Let $f:0,1^Sto[0,1]$ be a continuous function and suppose $f(sigma)=1$ for some $sigmain0,1^S$. I need to show that $f(tau)=1$ for some $tauin0,1^Ssetminussigma$.
My first instinct was to take note of the cardinalities. Since $S$ is uncountable, $0,1^S$ has a cardinality of at least $2^aleph_1$ whereas $[0,1]$ has a cardinality of only $2^aleph_0$, so $f$ cannot be injective. Furthermore, every nonempty basis element in the product topology is a product of subsets of $0,1$ uncountably many of which must be $0,1$. So, every nonempty basis element, and therefore every nonempty open set in $0,1^S$, must have a cardinality of at least $2^aleph_1$.
Since $f$ is continuous, $(a,1]$ is open in $[0,1]$, and $f(sigma)=1$, $ f^-1(a,1]$ must have a cardinality of at least $2^aleph_1$, so $f$ cannot be one-to-one on any interval containing $1$. In fact, for any interval $(a,1]$, there must exist $bin(a,1]$ such that $f^-1(b)$ is uncountable. My problem is that I have no idea how to use this to conclude that $f^-1(1)$ contains more than just $sigma$.
Another potentially useful thing is that $0,1^S$ is compact since it is a product of compact sets which means the Extreme Value Theorem applies here.
general-topology
asked Jul 24 at 11:12
Julian Benali
25712
add a comment |Â
up vote
6
down vote
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Let $S$ be an uncountable set and consider the space $0,1^S$ under the product topology where $0,1$ is discrete. Let $f:0,1^Sto[0,1]$ be a continuous function and suppose $f(sigma)=1$ for some $sigmain0,1^S$. I need to show that $f(tau)=1$ for some $tauin0,1^Ssetminussigma$.
My first instinct was to take note of the cardinalities. Since $S$ is uncountable, $0,1^S$ has a cardinality of at least $2^aleph_1$ whereas $[0,1]$ has a cardinality of only $2^aleph_0$, so $f$ cannot be injective. Furthermore, every nonempty basis element in the product topology is a product of subsets of $0,1$ uncountably many of which must be $0,1$. So, every nonempty basis element, and therefore every nonempty open set in $0,1^S$, must have a cardinality of at least $2^aleph_1$.
Since $f$ is continuous, $(a,1]$ is open in $[0,1]$, and $f(sigma)=1$, $ f^-1(a,1]$ must have a cardinality of at least $2^aleph_1$, so $f$ cannot be one-to-one on any interval containing $1$. In fact, for any interval $(a,1]$, there must exist $bin(a,1]$ such that $f^-1(b)$ is uncountable. My problem is that I have no idea how to use this to conclude that $f^-1(1)$ contains more than just $sigma$.
Another potentially useful thing is that $0,1^S$ is compact since it is a product of compact sets which means the Extreme Value Theorem applies here.
general-topology
asked Jul 24 at 11:12
Julian Benali
25712
1
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $S$ be an uncountable set and consider the space $0,1^S$ under the product topology where $0,1$ is discrete. Let $f:0,1^Sto[0,1]$ be a continuous function and suppose $f(sigma)=1$ for some $sigmain0,1^S$. I need to show that $f(tau)=1$ for some $tauin0,1^Ssetminussigma$.
My first instinct was to take note of the cardinalities. Since $S$ is uncountable, $0,1^S$ has a cardinality of at least $2^aleph_1$ whereas $[0,1]$ has a cardinality of only $2^aleph_0$, so $f$ cannot be injective. Furthermore, every nonempty basis element in the product topology is a product of subsets of $0,1$ uncountably many of which must be $0,1$. So, every nonempty basis element, and therefore every nonempty open set in $0,1^S$, must have a cardinality of at least $2^aleph_1$.
Since $f$ is continuous, $(a,1]$ is open in $[0,1]$, and $f(sigma)=1$, $ f^-1(a,1]$ must have a cardinality of at least $2^aleph_1$, so $f$ cannot be one-to-one on any interval containing $1$. In fact, for any interval $(a,1]$, there must exist $bin(a,1]$ such that $f^-1(b)$ is uncountable. My problem is that I have no idea how to use this to conclude that $f^-1(1)$ contains more than just $sigma$.
Another potentially useful thing is that $0,1^S$ is compact since it is a product of compact sets which means the Extreme Value Theorem applies here.
general-topology
asked Jul 24 at 11:12
Julian Benali
25712
Let $S$ be an uncountable set and consider the space $0,1^S$ under the product topology where $0,1$ is discrete. Let $f:0,1^Sto[0,1]$ be a continuous function and suppose $f(sigma)=1$ for some $sigmain0,1^S$. I need to show that $f(tau)=1$ for some $tauin0,1^Ssetminussigma$.
My first instinct was to take note of the cardinalities. Since $S$ is uncountable, $0,1^S$ has a cardinality of at least $2^aleph_1$ whereas $[0,1]$ has a cardinality of only $2^aleph_0$, so $f$ cannot be injective. Furthermore, every nonempty basis element in the product topology is a product of subsets of $0,1$ uncountably many of which must be $0,1$. So, every nonempty basis element, and therefore every nonempty open set in $0,1^S$, must have a cardinality of at least $2^aleph_1$.
Since $f$ is continuous, $(a,1]$ is open in $[0,1]$, and $f(sigma)=1$, $ f^-1(a,1]$ must have a cardinality of at least $2^aleph_1$, so $f$ cannot be one-to-one on any interval containing $1$. In fact, for any interval $(a,1]$, there must exist $bin(a,1]$ such that $f^-1(b)$ is uncountable. My problem is that I have no idea how to use this to conclude that $f^-1(1)$ contains more than just $sigma$.
Another potentially useful thing is that $0,1^S$ is compact since it is a product of compact sets which means the Extreme Value Theorem applies here.
general-topology
asked Jul 24 at 11:12
Julian Benali
25712
asked Jul 24 at 11:12
Julian Benali
25712
asked Jul 24 at 11:12
Julian Benali
25712
asked Jul 24 at 11:12
Julian Benali
25712
25712
1
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53
add a comment |Â
1
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53
1
1
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53
add a comment |Â
1 Answer
1
active
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votes
up vote
5
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accepted
Note that $f^-1(1) = cap_n>0, f^-1left((1-n^-1, 1]right)$. In particular $sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $Tsubset S$ such that $f(tau) = f(sigma)$ if $tau$ and $sigma$ coincide on $T$.
answered Jul 24 at 11:48
WimC
23.7k22860
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Note that $f^-1(1) = cap_n>0, f^-1left((1-n^-1, 1]right)$. In particular $sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $Tsubset S$ such that $f(tau) = f(sigma)$ if $tau$ and $sigma$ coincide on $T$.
answered Jul 24 at 11:48
WimC
23.7k22860
add a comment |Â
up vote
5
down vote
accepted
Note that $f^-1(1) = cap_n>0, f^-1left((1-n^-1, 1]right)$. In particular $sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $Tsubset S$ such that $f(tau) = f(sigma)$ if $tau$ and $sigma$ coincide on $T$.
answered Jul 24 at 11:48
WimC
23.7k22860
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Note that $f^-1(1) = cap_n>0, f^-1left((1-n^-1, 1]right)$. In particular $sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $Tsubset S$ such that $f(tau) = f(sigma)$ if $tau$ and $sigma$ coincide on $T$.
answered Jul 24 at 11:48
WimC
23.7k22860
Note that $f^-1(1) = cap_n>0, f^-1left((1-n^-1, 1]right)$. In particular $sigma$ is contained in a countable intersection of basis open sets. Conclude that there is an enumerable set $Tsubset S$ such that $f(tau) = f(sigma)$ if $tau$ and $sigma$ coincide on $T$.
answered Jul 24 at 11:48
WimC
23.7k22860
edited Jul 24 at 13:29
answered Jul 24 at 11:48
WimC
23.7k22860
answered Jul 24 at 11:48
WimC
23.7k22860
answered Jul 24 at 11:48
WimC
23.7k22860
23.7k22860
add a comment |Â
add a comment |Â
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1
Whether $2^aleph_0<2^aleph_1$ is independent of the usual axioms of set theory (ZFC). If you weren't given that $f$ is continuous, it might, for all we know, be injective.
– Andreas Blass
Jul 24 at 12:52
Really? I realize that whether $aleph_1=2^aleph_0$ depends on the CH, but I was under the impression that $a<b$ implies $2^a<2^b$ was provable within ZFC. My bad!
– Julian Benali
Jul 24 at 13:53