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How to define limit operations in general topological spaces? Are nets able to do this?
is a net stronger than a transfinite sequence for characterizing topology?Permitted value of epsilon in Discrete Metric SpaceWhy are topological spaces interesting to study?Define $f(y)=d(x_0,y)$, prove that $f$ is continuous.Finite point set has limit points for general topological spaces?Is the uniform limit of continuous functions continuous for topological spaces?Can we define the concept of limit without a topology?Equivalent definition of limit of a function (Reference request)Can limits be defined in a more algebraic way, instead of using the completely analytic $delta$-$epsilon$ definition?Is Wikipedia correct about bounded sets?
$begingroup$
I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.
For $f:mathbbRto mathbbR$,
$
lim_xto x_0f(x)=a
$
means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.
Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?
real-analysis general-topology
$endgroup$
add a comment |
$begingroup$
I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.
For $f:mathbbRto mathbbR$,
$
lim_xto x_0f(x)=a
$
means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.
Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?
real-analysis general-topology
$endgroup$
add a comment |
$begingroup$
I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.
For $f:mathbbRto mathbbR$,
$
lim_xto x_0f(x)=a
$
means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.
Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?
real-analysis general-topology
$endgroup$
I was always under the impression that in order to take a limit, I need to have a metric defined on my underlying space.
For $f:mathbbRto mathbbR$,
$
lim_xto x_0f(x)=a
$
means that for all $epsilon >0 $ there exists a $delta(epsilon)>0$ such that $|f(x)-a|<epsilon$ whenever $0< |x-x_0|<delta$. The notion of a limit uses the underlying metric $|cdot|$ of $mathbbR$.
Is there a consistent way of dispensing with the metric and still define a limit operation in some topological space? Are nets able to do this?
real-analysis general-topology
real-analysis general-topology
edited Mar 17 at 14:17
YuiTo Cheng
2,0612637
2,0612637
asked Mar 17 at 11:12
EEEBEEEB
53138
53138
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The notion of limit is well-defined for any topological space, even non-metric ones.
Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.
But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.
$endgroup$
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
add a comment |
$begingroup$
The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.
Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.
$endgroup$
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
add a comment |
$begingroup$
A net is a function from an directed set $(I, le)$ (say) to a space $X$.
$f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.
The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.
I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.
If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The notion of limit is well-defined for any topological space, even non-metric ones.
Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.
But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.
$endgroup$
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
add a comment |
$begingroup$
The notion of limit is well-defined for any topological space, even non-metric ones.
Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.
But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.
$endgroup$
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
add a comment |
$begingroup$
The notion of limit is well-defined for any topological space, even non-metric ones.
Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.
But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.
$endgroup$
The notion of limit is well-defined for any topological space, even non-metric ones.
Here is the correct definition: let $f : X rightarrow Y$ be a function between two topological spaces. We say that the limit of $f$ at a point $x in X$ is the point $y in Y$ if for all neighborhoods $N$ of $y$ in $Y$, there exists a neighborhood $M$ of $x$ in $X$ such that $f(M) subset N$.
But note that the sequential characterization of limit ($f$ tends to $y$ in $x$ iff for every sequence $(x_n) rightarrow x$, one has $f(x_n) rightarrow y$) is not true in a general topological space. It is true if $X$ is metrizable.
edited Mar 17 at 14:46
psmears
71149
71149
answered Mar 17 at 13:09
TheSilverDoeTheSilverDoe
4,358114
4,358114
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
add a comment |
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
4
4
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
Perhaps it's worth mentioning that the limit might fail to be unique, in non-Hausdorff spaces.
$endgroup$
– chi
Mar 17 at 19:05
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
$begingroup$
@chi Yes, thank you for this comment ! Indeed the precision can be usefull.
$endgroup$
– TheSilverDoe
Mar 17 at 19:07
add a comment |
$begingroup$
The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.
Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.
$endgroup$
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
add a comment |
$begingroup$
The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.
Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.
$endgroup$
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
add a comment |
$begingroup$
The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.
Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.
$endgroup$
The general definition of continuity is as follows: let $X$ and $Y$ be topological spaces and $f:X to Y$ be a map. $F$ is continuous if the inverse image of any open set is open. $f$ is continuous at a point $x$ iff for every open set $V$ containing $f(x)$ there exists an open set $U$ containing $x$ such that $f(U) subset V$.
Continuity of $f$ is equivalent to the following: whenever a net $(x_i)_i in I$ converges to some point $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$. Similarly, $f$ is continuous at a point $x$ iff whenever a net $(x_i)_i in I$ converges to $x$ we have $x$ we have $(f(x_i))_i in I$ converges to $f(x)$.
answered Mar 17 at 13:11
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
68.6k53169
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
add a comment |
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
1
1
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
$begingroup$
But the question was about limits, not continuity.
$endgroup$
– Paul Sinclair
2 days ago
add a comment |
$begingroup$
A net is a function from an directed set $(I, le)$ (say) to a space $X$.
$f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.
The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.
I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.
If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.
$endgroup$
add a comment |
$begingroup$
A net is a function from an directed set $(I, le)$ (say) to a space $X$.
$f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.
The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.
I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.
If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.
$endgroup$
add a comment |
$begingroup$
A net is a function from an directed set $(I, le)$ (say) to a space $X$.
$f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.
The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.
I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.
If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.
$endgroup$
A net is a function from an directed set $(I, le)$ (say) to a space $X$.
$f: I to X$ converges to $x$ iff for every open set $O$ that contains $x$ there exists some $i_0 in I$ (depending on $O$, in general) such that for all $i in I, i ge i_0$ we know that $f(i) in O$.
The point $f(i)$ is often denoted by subscript: $x_i$. This subscript an be any member of the directed set $I$. A sequence is the special case where $I=mathbbN, le)$ (in its standard order). The definition is non-metric in that we use open sets containing $x$ (not open balls) but for metric spaces it suffices to check this for open balls $B(x,varepsilon), varepsilon>0$ as these form a local base at $x$.
I think that $lim_x to a f(x)$ can be defined by considering all nets $n$ on $Xsetminus a$ that converge to $a$, and if all those nets have the property that $f circ n$ is a net in $Y$ converging to the same $b in Y$, then this $b$ is called the limit of $f$ as $x$ tends to $a$.
If $X$ has a topology induced from a metric, e.g. then we can restrict ourselves to sequences instead of general nets in the above characterisation.
edited Mar 17 at 16:44
answered Mar 17 at 13:40
Henno BrandsmaHenno Brandsma
113k348123
113k348123
add a comment |
add a comment |
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