Find all sequences, which convergence to $x_1 = 0$ and $x_2= frac{1}{3}$
We have basis of topology $tau = {{-r,r}: r>0} cup mathbb{R}$.
I want to find all sequences, which convergence to $x_1 = 0$ and $x_2= frac{1}{3}$.
The neighbourhood of $x_1 = 0$ is $mathbb{R}$. So it can be all sequences, which equal $0$?
The neighbourhood of $x_1 = frac{1}{3}$ is ${-frac{1}{3}, frac{1}{3}}$. So, here its all sequences, which belong to ${-frac{1}{3}, frac{1}{3}}$?
general-topology
edited Nov 24 at 13:11
Robert Z
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
add a comment |
We have basis of topology $tau = {{-r,r}: r>0} cup mathbb{R}$.
I want to find all sequences, which convergence to $x_1 = 0$ and $x_2= frac{1}{3}$.
The neighbourhood of $x_1 = 0$ is $mathbb{R}$. So it can be all sequences, which equal $0$?
The neighbourhood of $x_1 = frac{1}{3}$ is ${-frac{1}{3}, frac{1}{3}}$. So, here its all sequences, which belong to ${-frac{1}{3}, frac{1}{3}}$?
general-topology
edited Nov 24 at 13:11
Robert Z
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
1
Why a downvote?
– Robert Z
Nov 24 at 13:12
add a comment |
We have basis of topology $tau = {{-r,r}: r>0} cup mathbb{R}$.
I want to find all sequences, which convergence to $x_1 = 0$ and $x_2= frac{1}{3}$.
The neighbourhood of $x_1 = 0$ is $mathbb{R}$. So it can be all sequences, which equal $0$?
The neighbourhood of $x_1 = frac{1}{3}$ is ${-frac{1}{3}, frac{1}{3}}$. So, here its all sequences, which belong to ${-frac{1}{3}, frac{1}{3}}$?
general-topology
edited Nov 24 at 13:11
Robert Z
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
We have basis of topology $tau = {{-r,r}: r>0} cup mathbb{R}$.
I want to find all sequences, which convergence to $x_1 = 0$ and $x_2= frac{1}{3}$.
The neighbourhood of $x_1 = 0$ is $mathbb{R}$. So it can be all sequences, which equal $0$?
The neighbourhood of $x_1 = frac{1}{3}$ is ${-frac{1}{3}, frac{1}{3}}$. So, here its all sequences, which belong to ${-frac{1}{3}, frac{1}{3}}$?
general-topology
general-topology
edited Nov 24 at 13:11
Robert Z
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
edited Nov 24 at 13:11
Robert Z
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
edited Nov 24 at 13:11
Robert Z
93.2k1061132
edited Nov 24 at 13:11
Robert Z
93.2k1061132
edited Nov 24 at 13:11
Robert Z
93.2k1061132
93.2k1061132
asked Nov 24 at 12:59
Aleksandra
565
asked Nov 24 at 12:59
Aleksandra
565
asked Nov 24 at 12:59
Aleksandra
565
565
Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
1
Why a downvote?
– Robert Z
Nov 24 at 13:12
add a comment |
Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
1
Why a downvote?
– Robert Z
Nov 24 at 13:12
Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
1
1
Why a downvote?
– Robert Z
Nov 24 at 13:12
Why a downvote?
– Robert Z
Nov 24 at 13:12
add a comment |
1 Answer
1
active
oldest
votes
Recall that a sequence $(x_n)_n$ in a topology space $(X,tau)$ converges to a point $x in X$, if for each neighbourhood $U$ of $x$, there exists an $N in mathbb{N}$ such that for all $n geq N$ we have that $x_n in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood ${-1/3,1/3}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in ${-1/3,1/3}$.
Note that $(X,tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.
answered Nov 24 at 13:10
Robert Z
93.2k1061132
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Recall that a sequence $(x_n)_n$ in a topology space $(X,tau)$ converges to a point $x in X$, if for each neighbourhood $U$ of $x$, there exists an $N in mathbb{N}$ such that for all $n geq N$ we have that $x_n in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood ${-1/3,1/3}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in ${-1/3,1/3}$.
Note that $(X,tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.
answered Nov 24 at 13:10
Robert Z
93.2k1061132
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
add a comment |
Recall that a sequence $(x_n)_n$ in a topology space $(X,tau)$ converges to a point $x in X$, if for each neighbourhood $U$ of $x$, there exists an $N in mathbb{N}$ such that for all $n geq N$ we have that $x_n in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood ${-1/3,1/3}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in ${-1/3,1/3}$.
Note that $(X,tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.
answered Nov 24 at 13:10
Robert Z
93.2k1061132
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
add a comment |
Recall that a sequence $(x_n)_n$ in a topology space $(X,tau)$ converges to a point $x in X$, if for each neighbourhood $U$ of $x$, there exists an $N in mathbb{N}$ such that for all $n geq N$ we have that $x_n in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood ${-1/3,1/3}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in ${-1/3,1/3}$.
Note that $(X,tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.
answered Nov 24 at 13:10
Robert Z
93.2k1061132
Recall that a sequence $(x_n)_n$ in a topology space $(X,tau)$ converges to a point $x in X$, if for each neighbourhood $U$ of $x$, there exists an $N in mathbb{N}$ such that for all $n geq N$ we have that $x_n in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood ${-1/3,1/3}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in ${-1/3,1/3}$.
Note that $(X,tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.
answered Nov 24 at 13:10
Robert Z
93.2k1061132
edited Nov 24 at 13:36
answered Nov 24 at 13:10
Robert Z
93.2k1061132
answered Nov 24 at 13:10
Robert Z
93.2k1061132
answered Nov 24 at 13:10
Robert Z
93.2k1061132
93.2k1061132
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
add a comment |
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
I edited my answer. Is it clear now?
– Robert Z
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
Yes, now its clear!
– Aleksandra
Nov 24 at 13:29
add a comment |
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Do you mean the interval $(-r,r)$ instead of ${-r,r} $?
– Robert Z
Nov 24 at 13:02
@RobertZ No, it is not interval. Its a set, which consist of 2 numbers.
– Aleksandra
Nov 24 at 13:06
1
Why a downvote?
– Robert Z
Nov 24 at 13:12