Revisions to The difference between a sequential space and a space with countable tightness

tightened it up a bit, fixed the definitions of bases

Source Link

edited Feb 4, 2011 at 19:49

703
4
8

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

added 64 characters in body

Source Link

edited Feb 4, 2011 at 16:56

Apollo

703
4
8

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N\}$$A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N\in\omega$$M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N\}$$A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N\in\omega$$M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N\}$ for $M,N\in\omega$ - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N\}$ for $M,N\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N_m\}$ for $M,N_m\in\omega$ (ie cofinitely many elements of cofinitely many fibers) - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N_m\}$ for $M,N_m\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Source Link

created Feb 2, 2011 at 23:20

Apollo

703
4
8

Some examples to expand Andreas' answer might be of interest: (it's too much to fit in a comment so I'm adding it as an answer, though I think Andreas' response is great)

any metric space will be Frechet-Urysohn (choose $x_n$ in $A$ within $1/n$ of $p$); (more generally, any first-countable space is F-U: just choose $x_n$ in the intersection of $A$ and $U_n$ where $\{U_n\}_n$ is a countable base at the limit point);
a sequential but not Frechet-Urysohn space is given by taking $((\omega+1)\times\omega)\cup\{*\}$ where each copy of $\omega+1$ has the usual topology and a base for $*$ consists of sets $A_{m,n}=\{(m,n)|m>M,n>N\}$ for $M,N\in\omega$ - then $*$ is in the closure of $\omega\times\omega$ but is not the limit of any sequence of points in $\omega\times\omega$; however, it is the limit of the sequence $x_n=(\omega,n)$ and each $x_n$ is the limit of a sequence of points from $\omega\times\omega$;
a countably-tight but not sequential space could be given by taking $(\omega\times\omega)\cup\{*\}$ where all points $(m,n)$ are open and a base for $*$ consist of sets $A_{M,N}=\{(m,n)|m>M, n>N\}$ for $M,N\in\omega$ - this space is trivially countably tight (it's countable) but is not sequential: $*$ is not the limit of any sequence in $\omega\times\omega$ (since we can always exclude a cofinal subsequence of any putative sequence converging to $*$);
finally a non-countably-tight space is given by $\omega_1+1$ with the usual topology: $\omega_1$ (as a point) is in the closure of $\omega_1$ (as a set) but any countable subset of $\omega_1$ has bounded (countable) closure.

Stack Exchange Network

Return to Answer