Mixing
Two Hausdorff topologies on the same set with the same convergent nets
$begingroup$
Let $X$ be a set and $mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{alpha})_{alpha}$ in $X$ converges with respect to $mathcal{T_1}$ iff it converges with respect to $mathcal{T_2}$ (but to a possibly different point). Does it follow that $mathcal{T_1=T_2}$?
If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set ${a,b}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?
general-topology nets
asked Jan 31 at 13:04
ColescuColescu
3,23511136
$endgroup$
add a comment |
$begingroup$
Let $X$ be a set and $mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{alpha})_{alpha}$ in $X$ converges with respect to $mathcal{T_1}$ iff it converges with respect to $mathcal{T_2}$ (but to a possibly different point). Does it follow that $mathcal{T_1=T_2}$?
If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set ${a,b}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?
general-topology nets
asked Jan 31 at 13:04
ColescuColescu
3,23511136
$endgroup$
1
$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22
add a comment |
$begingroup$
Let $X$ be a set and $mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{alpha})_{alpha}$ in $X$ converges with respect to $mathcal{T_1}$ iff it converges with respect to $mathcal{T_2}$ (but to a possibly different point). Does it follow that $mathcal{T_1=T_2}$?
If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set ${a,b}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?
general-topology nets
asked Jan 31 at 13:04
ColescuColescu
3,23511136
$endgroup$
Let $X$ be a set and $mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{alpha})_{alpha}$ in $X$ converges with respect to $mathcal{T_1}$ iff it converges with respect to $mathcal{T_2}$ (but to a possibly different point). Does it follow that $mathcal{T_1=T_2}$?
If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set ${a,b}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?
general-topology nets
general-topology nets
asked Jan 31 at 13:04
ColescuColescu
3,23511136
asked Jan 31 at 13:04
ColescuColescu
3,23511136
asked Jan 31 at 13:04
ColescuColescu
3,23511136
asked Jan 31 at 13:04
ColescuColescu
3,23511136
asked Jan 31 at 13:04
ColescuColescu
3,23511136
3,23511136
1
$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22
add a comment |
1
$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22
1
1
$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22
$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $(x_{alpha})_{alphain A}$ be a net converging in to $x$ in $mathcal{T}_{1}$ and to $y$ in $mathcal{T}_{2}$. We define
$$hat{A}=Atimes{1,2}$$
and we say $(alpha,i)leq(beta,j)$ iff $alphalneqbeta$ or $alpha=beta$ and $ileq j$. Note that $hat{A}$, $Atimes{1}$ and $Atimes{2}$ are directed sets. We define the net $(y_{(alpha,i)})_{hat{A}}$ by
$$y_{(alpha,i)}=begin{cases}x_{alpha}&text{ if }i=1\ y&text{ if }i=2end{cases}.$$
Clearly $(y_{(alpha,i)})_{hat{A}}$ converges to $y$ in $mathcal{T}_{2}$, but in $mathcal{T}_{1}$ the subnet $(y_{(alpha,i)})_{Atimes{1}}$ converges to $x$ and the subnet $(y_{(alpha,i)})_{Atimes{2}}$ converges to $y$.
Thus a net converges in $mathcal{T}_{1}$ if and only if it converges in $mathcal{T}_{2}$ convergent nets converge to the same point.
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
$endgroup$
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $(x_{alpha})_{alphain A}$ be a net converging in to $x$ in $mathcal{T}_{1}$ and to $y$ in $mathcal{T}_{2}$. We define
$$hat{A}=Atimes{1,2}$$
and we say $(alpha,i)leq(beta,j)$ iff $alphalneqbeta$ or $alpha=beta$ and $ileq j$. Note that $hat{A}$, $Atimes{1}$ and $Atimes{2}$ are directed sets. We define the net $(y_{(alpha,i)})_{hat{A}}$ by
$$y_{(alpha,i)}=begin{cases}x_{alpha}&text{ if }i=1\ y&text{ if }i=2end{cases}.$$
Clearly $(y_{(alpha,i)})_{hat{A}}$ converges to $y$ in $mathcal{T}_{2}$, but in $mathcal{T}_{1}$ the subnet $(y_{(alpha,i)})_{Atimes{1}}$ converges to $x$ and the subnet $(y_{(alpha,i)})_{Atimes{2}}$ converges to $y$.
Thus a net converges in $mathcal{T}_{1}$ if and only if it converges in $mathcal{T}_{2}$ convergent nets converge to the same point.
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
$endgroup$
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
add a comment |
$begingroup$
Let $(x_{alpha})_{alphain A}$ be a net converging in to $x$ in $mathcal{T}_{1}$ and to $y$ in $mathcal{T}_{2}$. We define
$$hat{A}=Atimes{1,2}$$
and we say $(alpha,i)leq(beta,j)$ iff $alphalneqbeta$ or $alpha=beta$ and $ileq j$. Note that $hat{A}$, $Atimes{1}$ and $Atimes{2}$ are directed sets. We define the net $(y_{(alpha,i)})_{hat{A}}$ by
$$y_{(alpha,i)}=begin{cases}x_{alpha}&text{ if }i=1\ y&text{ if }i=2end{cases}.$$
Clearly $(y_{(alpha,i)})_{hat{A}}$ converges to $y$ in $mathcal{T}_{2}$, but in $mathcal{T}_{1}$ the subnet $(y_{(alpha,i)})_{Atimes{1}}$ converges to $x$ and the subnet $(y_{(alpha,i)})_{Atimes{2}}$ converges to $y$.
Thus a net converges in $mathcal{T}_{1}$ if and only if it converges in $mathcal{T}_{2}$ convergent nets converge to the same point.
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
$endgroup$
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
add a comment |
$begingroup$
Let $(x_{alpha})_{alphain A}$ be a net converging in to $x$ in $mathcal{T}_{1}$ and to $y$ in $mathcal{T}_{2}$. We define
$$hat{A}=Atimes{1,2}$$
and we say $(alpha,i)leq(beta,j)$ iff $alphalneqbeta$ or $alpha=beta$ and $ileq j$. Note that $hat{A}$, $Atimes{1}$ and $Atimes{2}$ are directed sets. We define the net $(y_{(alpha,i)})_{hat{A}}$ by
$$y_{(alpha,i)}=begin{cases}x_{alpha}&text{ if }i=1\ y&text{ if }i=2end{cases}.$$
Clearly $(y_{(alpha,i)})_{hat{A}}$ converges to $y$ in $mathcal{T}_{2}$, but in $mathcal{T}_{1}$ the subnet $(y_{(alpha,i)})_{Atimes{1}}$ converges to $x$ and the subnet $(y_{(alpha,i)})_{Atimes{2}}$ converges to $y$.
Thus a net converges in $mathcal{T}_{1}$ if and only if it converges in $mathcal{T}_{2}$ convergent nets converge to the same point.
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
$endgroup$
Let $(x_{alpha})_{alphain A}$ be a net converging in to $x$ in $mathcal{T}_{1}$ and to $y$ in $mathcal{T}_{2}$. We define
$$hat{A}=Atimes{1,2}$$
and we say $(alpha,i)leq(beta,j)$ iff $alphalneqbeta$ or $alpha=beta$ and $ileq j$. Note that $hat{A}$, $Atimes{1}$ and $Atimes{2}$ are directed sets. We define the net $(y_{(alpha,i)})_{hat{A}}$ by
$$y_{(alpha,i)}=begin{cases}x_{alpha}&text{ if }i=1\ y&text{ if }i=2end{cases}.$$
Clearly $(y_{(alpha,i)})_{hat{A}}$ converges to $y$ in $mathcal{T}_{2}$, but in $mathcal{T}_{1}$ the subnet $(y_{(alpha,i)})_{Atimes{1}}$ converges to $x$ and the subnet $(y_{(alpha,i)})_{Atimes{2}}$ converges to $y$.
Thus a net converges in $mathcal{T}_{1}$ if and only if it converges in $mathcal{T}_{2}$ convergent nets converge to the same point.
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
answered Jan 31 at 13:30
Floris ClaassensFloris Claassens
1,33229
1,33229
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
add a comment |
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
$Atimes{1}$ is cofinal in $hat{A}$. Let $(alpha,i)inhat{A}$, there exists a $betain A$ such that $alphaleqbeta$ as $A$ is a directed set. Then $(beta,1)geq(alpha,i)$.
$endgroup$
– Floris Claassens
Jan 31 at 13:38
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
So $(alpha,1)leq(alpha,2)$ and $(alpha,1)geq(alpha,2)$? In that case $(y_{(alpha,i)})_{hat{A}}$ may not converge to $y$...
$endgroup$
– Colescu
Jan 31 at 13:42
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
Good point, no it does not, I was a bit sloppy in writing that comment. You need to take a $betain A$ such that $alphalneqbeta$. Note that if such a $beta$ does not exists, then $alpha$ is an upper bound for $A$ in which case $x_{alpha}=x=y$.
$endgroup$
– Floris Claassens
Jan 31 at 13:52
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
$begingroup$
I see, so it works! Thanks for the elegant solutoin. :)
$endgroup$
– Colescu
Jan 31 at 13:56
add a comment |
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$begingroup$
A convergent sequence $x_1,x_2,x_3,dots$ converges to the point $x$ if and only if the sequence $x_1,x,x_2,x,x_3,x,cdots$ is convergent. See if you can to something similar with nets.
$endgroup$
– GEdgar
Jan 31 at 13:22