Closure and accumulation point in $mathbb R^2$ The 2019 Stack Overflow Developer Survey Results Are InAccumulation point in $mathbb R^k$Question regarding infinite subsets of R and accumulation pointsProperties of ClosureAccumulation points in Caccumulation point of the set $(-1,0]$.What is the closure of $(0,1)$ in $mathbbR_k$?Concerning the closure of a set in this topologyIs every accumulation point of the set of values of sequence $x_n$ also the accumulation point of the sequence $x_n$.accumulation point of infinite compact subset of $mathbbR^2$Set of accumulation points
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Closure and accumulation point in $mathbb R^2$
The 2019 Stack Overflow Developer Survey Results Are InAccumulation point in $mathbb R^k$Question regarding infinite subsets of R and accumulation pointsProperties of ClosureAccumulation points in Caccumulation point of the set $(-1,0]$.What is the closure of $(0,1)$ in $mathbbR_k$?Concerning the closure of a set in this topologyIs every accumulation point of the set of values of sequence $x_n$ also the accumulation point of the sequence $x_n$.accumulation point of infinite compact subset of $mathbbR^2$Set of accumulation points
$begingroup$
I gave a topology defined on $mathbbR^2$ by it basis
$$O_m=(x,y)inmathbbR^2, max(x,y)leq m$$
where $min mathbbN$.
The question is to find the closure and the set of accumulation points of
$$A=(x,y)in mathbbR^2, x^2+y^2-6x+5leq 0$$
I found
$$cl(A)= mathbbR^2- O_0$$
and
$$A'=cl(A)-(1,0)$$
is it right ? thank you
general-topology
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
$endgroup$
add a comment |
$begingroup$
I gave a topology defined on $mathbbR^2$ by it basis
$$O_m=(x,y)inmathbbR^2, max(x,y)leq m$$
where $min mathbbN$.
The question is to find the closure and the set of accumulation points of
$$A=(x,y)in mathbbR^2, x^2+y^2-6x+5leq 0$$
I found
$$cl(A)= mathbbR^2- O_0$$
and
$$A'=cl(A)-(1,0)$$
is it right ? thank you
general-topology
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
$endgroup$
add a comment |
$begingroup$
I gave a topology defined on $mathbbR^2$ by it basis
$$O_m=(x,y)inmathbbR^2, max(x,y)leq m$$
where $min mathbbN$.
The question is to find the closure and the set of accumulation points of
$$A=(x,y)in mathbbR^2, x^2+y^2-6x+5leq 0$$
I found
$$cl(A)= mathbbR^2- O_0$$
and
$$A'=cl(A)-(1,0)$$
is it right ? thank you
general-topology
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
$endgroup$
I gave a topology defined on $mathbbR^2$ by it basis
$$O_m=(x,y)inmathbbR^2, max(x,y)leq m$$
where $min mathbbN$.
The question is to find the closure and the set of accumulation points of
$$A=(x,y)in mathbbR^2, x^2+y^2-6x+5leq 0$$
I found
$$cl(A)= mathbbR^2- O_0$$
and
$$A'=cl(A)-(1,0)$$
is it right ? thank you
general-topology
general-topology
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
edited Apr 7 at 10:03
YuiTo Cheng
2,3694937
2,3694937
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
asked Apr 6 at 19:27
Poline SandraPoline Sandra
1117
1117
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not hard to see that $A$ is the set of all points distance $d le 2$ from
the point $(3,0)$.
Note that the only open sets contained in the complement of $A$ are $emptyset$, $...,O_-2,O_-1,O_0$ and hence the smallest closed set containing $A$ is $O_0^c$.
The set of accumulation points is more involved. I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Note that a point $(x,y)$ is only contained in the open sets $O_k,O_k+1,...$ where
$k=max(lceil x rceil, lceil y rceil)$.
Note that the open set $O_k$ intersects $A$ iff $k ge 1$.
In particular, the set of accumulation points is $O_0^c setminus (1,0)$.
Aside: Note that from the perspective of the topology, points in $O_k setminus O_k-1$ are 'indistinguishable' from each other. In particular, any point in $O_1 setminus O_0$ converges to any point in $O_1 setminus O_0$,
in particular $(1,0)$. Note that $(1,0)$ is an isolated point of $A$.
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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active
oldest
votes
$begingroup$
It is not hard to see that $A$ is the set of all points distance $d le 2$ from
the point $(3,0)$.
Note that the only open sets contained in the complement of $A$ are $emptyset$, $...,O_-2,O_-1,O_0$ and hence the smallest closed set containing $A$ is $O_0^c$.
The set of accumulation points is more involved. I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Note that a point $(x,y)$ is only contained in the open sets $O_k,O_k+1,...$ where
$k=max(lceil x rceil, lceil y rceil)$.
Note that the open set $O_k$ intersects $A$ iff $k ge 1$.
In particular, the set of accumulation points is $O_0^c setminus (1,0)$.
Aside: Note that from the perspective of the topology, points in $O_k setminus O_k-1$ are 'indistinguishable' from each other. In particular, any point in $O_1 setminus O_0$ converges to any point in $O_1 setminus O_0$,
in particular $(1,0)$. Note that $(1,0)$ is an isolated point of $A$.
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
It is not hard to see that $A$ is the set of all points distance $d le 2$ from
the point $(3,0)$.
Note that the only open sets contained in the complement of $A$ are $emptyset$, $...,O_-2,O_-1,O_0$ and hence the smallest closed set containing $A$ is $O_0^c$.
The set of accumulation points is more involved. I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Note that a point $(x,y)$ is only contained in the open sets $O_k,O_k+1,...$ where
$k=max(lceil x rceil, lceil y rceil)$.
Note that the open set $O_k$ intersects $A$ iff $k ge 1$.
In particular, the set of accumulation points is $O_0^c setminus (1,0)$.
Aside: Note that from the perspective of the topology, points in $O_k setminus O_k-1$ are 'indistinguishable' from each other. In particular, any point in $O_1 setminus O_0$ converges to any point in $O_1 setminus O_0$,
in particular $(1,0)$. Note that $(1,0)$ is an isolated point of $A$.
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
It is not hard to see that $A$ is the set of all points distance $d le 2$ from
the point $(3,0)$.
Note that the only open sets contained in the complement of $A$ are $emptyset$, $...,O_-2,O_-1,O_0$ and hence the smallest closed set containing $A$ is $O_0^c$.
The set of accumulation points is more involved. I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Note that a point $(x,y)$ is only contained in the open sets $O_k,O_k+1,...$ where
$k=max(lceil x rceil, lceil y rceil)$.
Note that the open set $O_k$ intersects $A$ iff $k ge 1$.
In particular, the set of accumulation points is $O_0^c setminus (1,0)$.
Aside: Note that from the perspective of the topology, points in $O_k setminus O_k-1$ are 'indistinguishable' from each other. In particular, any point in $O_1 setminus O_0$ converges to any point in $O_1 setminus O_0$,
in particular $(1,0)$. Note that $(1,0)$ is an isolated point of $A$.
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
$endgroup$
It is not hard to see that $A$ is the set of all points distance $d le 2$ from
the point $(3,0)$.
Note that the only open sets contained in the complement of $A$ are $emptyset$, $...,O_-2,O_-1,O_0$ and hence the smallest closed set containing $A$ is $O_0^c$.
The set of accumulation points is more involved. I am presuming that $p$ is an accumulation point iff any open set that contains $p$ also contains an element of $A$ different from $x$.
Note that a point $(x,y)$ is only contained in the open sets $O_k,O_k+1,...$ where
$k=max(lceil x rceil, lceil y rceil)$.
Note that the open set $O_k$ intersects $A$ iff $k ge 1$.
In particular, the set of accumulation points is $O_0^c setminus (1,0)$.
Aside: Note that from the perspective of the topology, points in $O_k setminus O_k-1$ are 'indistinguishable' from each other. In particular, any point in $O_1 setminus O_0$ converges to any point in $O_1 setminus O_0$,
in particular $(1,0)$. Note that $(1,0)$ is an isolated point of $A$.
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
edited Apr 7 at 6:58
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
answered Apr 7 at 0:23
copper.hatcopper.hat
128k561161
128k561161
add a comment |
add a comment |
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