What is the limit of $(1-frac1n)$ in this topology? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding limit points in lexicographic order topologyThe topology on $mathbbR$ with sub-basis consisting of all half open intervals $[a,b)$.Is this topology discrete?Specific instance of a general selection principle in topology.Compare this topology with the usual topologyLower Limit Topology?Convergence and topologyTopology over $mathbb N$.What is the adherent value of $(frac1n,1)$ in this topology?A set is compact in complement topology iff closed in standard topology
Can't figure this one out.. What is the missing box?
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What is the limit of $(1-frac1n)$ in this topology? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding limit points in lexicographic order topologyThe topology on $mathbbR$ with sub-basis consisting of all half open intervals $[a,b)$.Is this topology discrete?Specific instance of a general selection principle in topology.Compare this topology with the usual topologyLower Limit Topology?Convergence and topologyTopology over $mathbb N$.What is the adherent value of $(frac1n,1)$ in this topology?A set is compact in complement topology iff closed in standard topology
$begingroup$
On $E=[0,1[$ I consider the topology $$tau=[0,x[, xin [0,1]$$
How to find the limit of $u_n=1-frac1n$?
Thank you
general-topology
edited Mar 31 at 23:00
Paul Frost
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
$endgroup$
closed as off-topic by RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu Apr 1 at 4:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu
add a comment |
$begingroup$
On $E=[0,1[$ I consider the topology $$tau=[0,x[, xin [0,1]$$
How to find the limit of $u_n=1-frac1n$?
Thank you
general-topology
edited Mar 31 at 23:00
Paul Frost
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
$endgroup$
closed as off-topic by RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu Apr 1 at 4:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu
add a comment |
$begingroup$
On $E=[0,1[$ I consider the topology $$tau=[0,x[, xin [0,1]$$
How to find the limit of $u_n=1-frac1n$?
Thank you
general-topology
edited Mar 31 at 23:00
Paul Frost
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
$endgroup$
On $E=[0,1[$ I consider the topology $$tau=[0,x[, xin [0,1]$$
How to find the limit of $u_n=1-frac1n$?
Thank you
general-topology
general-topology
edited Mar 31 at 23:00
Paul Frost
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
edited Mar 31 at 23:00
Paul Frost
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
edited Mar 31 at 23:00
Paul Frost
12.7k31035
edited Mar 31 at 23:00
Paul Frost
12.7k31035
edited Mar 31 at 23:00
Paul Frost
12.7k31035
12.7k31035
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
asked Mar 31 at 18:14
Poline SandraPoline Sandra
1208
1208
closed as off-topic by RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu Apr 1 at 4:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu
closed as off-topic by RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu Apr 1 at 4:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Leucippus, GNUSupporter 8964民主女神 地下教會, Martin Argerami, Tianlalu
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: for any open set $[0,x[$, $0 < x < 1$: as $(1 - 1/n) > x$ for $n$ large enough, $(1 - 1/n)notin [0,x[$ for $n$ large enough...
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
add a comment |
$begingroup$
Put simply, you can't, because it doesn't exist.
Take any $xin E.$ By definition of $E,$ we know that $0le x<1,$ and so by definition of $tau,$ the open neighborhoods of $x$ will be of the form $[0,y[$ for some $yin]x,1].$ However, taking $y=fracx+12,$ we see that only finitely-many $u_n$ will be in the set $[0,y[,$ so $u_n$ can't converge to $x.$
On the other hand, if we'd had the set $E=[0,1]$ and had $tau=bigl[0,x[::xin[0,1]bigrcupE,$ then we would find that $1$ is the limit of $u_n.$
Or, with the original $E$ and $tau,$ if we'd considered $v_n=frac1n+1,$ we would find that $v_n$ converges to every point of $E$!
Let me see if I can expand on my approach above.
First of all, note that for any $xin E$ and any $varepsilon,$ the following are equivalent: $$xin[0,x+varepsilon[;text and ;[0,x+varepsilon[,intau\x<x+varepsilonle 1\0<varepsilonle 1-x\varepsilonin,]0,1-x]$$
Next, note that for any $ninBbb N,$ any $xin E,$ and any $varepsilonin,]0,1-x],$ the following are equivalent: $$u_nin[0,x+varepsilon[\0le1-frac1n<x+varepsilon\frac1nle 1<frac1n+x+varepsilon\1le n<1+(x+varepsilon)n\n<1+(x+varepsilon)n\n-(x+varepsilon)n<1\nbigl(1-(x+varepsilon)bigr)<1\n<frac11-(x+varepsilon)\1le n<frac11-(x+varepsilon)\1le n<leftlceilfrac11-(x+varepsilon)rightrceil\1le nleleftlceilfrac11-(x+varepsilon)rightrceil-1$$
Hence, we have in particular that there are only $leftlceilfrac11-(x+varepsilon)rightrceil-1$ values of $n$ for which $u_nin[0,x+varepsilon[.$ Since all neighborhoods of $x$ have the form $[0,x+varepsilon[$ for some $varepsilonin,]0,1-x],$ then regardless of $x,$ every neighborhood of $x$ has only finitely-many $u_n$ inside it.
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
$endgroup$
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
$endgroup$
– Cameron Buie
Mar 31 at 19:54
$begingroup$
I don't understand what you write
$endgroup$
– Poline Sandra
Mar 31 at 20:14
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
|
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: for any open set $[0,x[$, $0 < x < 1$: as $(1 - 1/n) > x$ for $n$ large enough, $(1 - 1/n)notin [0,x[$ for $n$ large enough...
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
add a comment |
$begingroup$
Hint: for any open set $[0,x[$, $0 < x < 1$: as $(1 - 1/n) > x$ for $n$ large enough, $(1 - 1/n)notin [0,x[$ for $n$ large enough...
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
add a comment |
$begingroup$
Hint: for any open set $[0,x[$, $0 < x < 1$: as $(1 - 1/n) > x$ for $n$ large enough, $(1 - 1/n)notin [0,x[$ for $n$ large enough...
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
$endgroup$
Hint: for any open set $[0,x[$, $0 < x < 1$: as $(1 - 1/n) > x$ for $n$ large enough, $(1 - 1/n)notin [0,x[$ for $n$ large enough...
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
answered Mar 31 at 18:23
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.5k42972
35.5k42972
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
add a comment |
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
how to prove that an infinite number is not in the open set ?
$endgroup$
– Poline Sandra
Mar 31 at 19:49
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
@PolineSandra, $(1−1/n)>xiff n > 1/(1-x)$.
$endgroup$
– Martín-Blas Pérez Pinilla
Mar 31 at 19:51
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
$begingroup$
for example why 0 is not a limit ? $u_nin [0,varepsilon[ iff n<1-varepsilon$ how to find the counter example such that 0 is not a limit
$endgroup$
– Poline Sandra
Mar 31 at 19:56
1
1
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
$begingroup$
@Poline: We can't have $0$ as a limit point, because $u_1=0$ is the only sequence element in $left[0,frac12right[.$
$endgroup$
– Cameron Buie
Mar 31 at 20:43
add a comment |
$begingroup$
Put simply, you can't, because it doesn't exist.
Take any $xin E.$ By definition of $E,$ we know that $0le x<1,$ and so by definition of $tau,$ the open neighborhoods of $x$ will be of the form $[0,y[$ for some $yin]x,1].$ However, taking $y=fracx+12,$ we see that only finitely-many $u_n$ will be in the set $[0,y[,$ so $u_n$ can't converge to $x.$
On the other hand, if we'd had the set $E=[0,1]$ and had $tau=bigl[0,x[::xin[0,1]bigrcupE,$ then we would find that $1$ is the limit of $u_n.$
Or, with the original $E$ and $tau,$ if we'd considered $v_n=frac1n+1,$ we would find that $v_n$ converges to every point of $E$!
Let me see if I can expand on my approach above.
First of all, note that for any $xin E$ and any $varepsilon,$ the following are equivalent: $$xin[0,x+varepsilon[;text and ;[0,x+varepsilon[,intau\x<x+varepsilonle 1\0<varepsilonle 1-x\varepsilonin,]0,1-x]$$
Next, note that for any $ninBbb N,$ any $xin E,$ and any $varepsilonin,]0,1-x],$ the following are equivalent: $$u_nin[0,x+varepsilon[\0le1-frac1n<x+varepsilon\frac1nle 1<frac1n+x+varepsilon\1le n<1+(x+varepsilon)n\n<1+(x+varepsilon)n\n-(x+varepsilon)n<1\nbigl(1-(x+varepsilon)bigr)<1\n<frac11-(x+varepsilon)\1le n<frac11-(x+varepsilon)\1le n<leftlceilfrac11-(x+varepsilon)rightrceil\1le nleleftlceilfrac11-(x+varepsilon)rightrceil-1$$
Hence, we have in particular that there are only $leftlceilfrac11-(x+varepsilon)rightrceil-1$ values of $n$ for which $u_nin[0,x+varepsilon[.$ Since all neighborhoods of $x$ have the form $[0,x+varepsilon[$ for some $varepsilonin,]0,1-x],$ then regardless of $x,$ every neighborhood of $x$ has only finitely-many $u_n$ inside it.
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
$endgroup$
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
$endgroup$
– Cameron Buie
Mar 31 at 19:54
$begingroup$
I don't understand what you write
$endgroup$
– Poline Sandra
Mar 31 at 20:14
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
|
show 5 more comments
$begingroup$
Put simply, you can't, because it doesn't exist.
Take any $xin E.$ By definition of $E,$ we know that $0le x<1,$ and so by definition of $tau,$ the open neighborhoods of $x$ will be of the form $[0,y[$ for some $yin]x,1].$ However, taking $y=fracx+12,$ we see that only finitely-many $u_n$ will be in the set $[0,y[,$ so $u_n$ can't converge to $x.$
On the other hand, if we'd had the set $E=[0,1]$ and had $tau=bigl[0,x[::xin[0,1]bigrcupE,$ then we would find that $1$ is the limit of $u_n.$
Or, with the original $E$ and $tau,$ if we'd considered $v_n=frac1n+1,$ we would find that $v_n$ converges to every point of $E$!
Let me see if I can expand on my approach above.
First of all, note that for any $xin E$ and any $varepsilon,$ the following are equivalent: $$xin[0,x+varepsilon[;text and ;[0,x+varepsilon[,intau\x<x+varepsilonle 1\0<varepsilonle 1-x\varepsilonin,]0,1-x]$$
Next, note that for any $ninBbb N,$ any $xin E,$ and any $varepsilonin,]0,1-x],$ the following are equivalent: $$u_nin[0,x+varepsilon[\0le1-frac1n<x+varepsilon\frac1nle 1<frac1n+x+varepsilon\1le n<1+(x+varepsilon)n\n<1+(x+varepsilon)n\n-(x+varepsilon)n<1\nbigl(1-(x+varepsilon)bigr)<1\n<frac11-(x+varepsilon)\1le n<frac11-(x+varepsilon)\1le n<leftlceilfrac11-(x+varepsilon)rightrceil\1le nleleftlceilfrac11-(x+varepsilon)rightrceil-1$$
Hence, we have in particular that there are only $leftlceilfrac11-(x+varepsilon)rightrceil-1$ values of $n$ for which $u_nin[0,x+varepsilon[.$ Since all neighborhoods of $x$ have the form $[0,x+varepsilon[$ for some $varepsilonin,]0,1-x],$ then regardless of $x,$ every neighborhood of $x$ has only finitely-many $u_n$ inside it.
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
$endgroup$
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
$endgroup$
– Cameron Buie
Mar 31 at 19:54
$begingroup$
I don't understand what you write
$endgroup$
– Poline Sandra
Mar 31 at 20:14
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
|
show 5 more comments
$begingroup$
Put simply, you can't, because it doesn't exist.
Take any $xin E.$ By definition of $E,$ we know that $0le x<1,$ and so by definition of $tau,$ the open neighborhoods of $x$ will be of the form $[0,y[$ for some $yin]x,1].$ However, taking $y=fracx+12,$ we see that only finitely-many $u_n$ will be in the set $[0,y[,$ so $u_n$ can't converge to $x.$
On the other hand, if we'd had the set $E=[0,1]$ and had $tau=bigl[0,x[::xin[0,1]bigrcupE,$ then we would find that $1$ is the limit of $u_n.$
Or, with the original $E$ and $tau,$ if we'd considered $v_n=frac1n+1,$ we would find that $v_n$ converges to every point of $E$!
Let me see if I can expand on my approach above.
First of all, note that for any $xin E$ and any $varepsilon,$ the following are equivalent: $$xin[0,x+varepsilon[;text and ;[0,x+varepsilon[,intau\x<x+varepsilonle 1\0<varepsilonle 1-x\varepsilonin,]0,1-x]$$
Next, note that for any $ninBbb N,$ any $xin E,$ and any $varepsilonin,]0,1-x],$ the following are equivalent: $$u_nin[0,x+varepsilon[\0le1-frac1n<x+varepsilon\frac1nle 1<frac1n+x+varepsilon\1le n<1+(x+varepsilon)n\n<1+(x+varepsilon)n\n-(x+varepsilon)n<1\nbigl(1-(x+varepsilon)bigr)<1\n<frac11-(x+varepsilon)\1le n<frac11-(x+varepsilon)\1le n<leftlceilfrac11-(x+varepsilon)rightrceil\1le nleleftlceilfrac11-(x+varepsilon)rightrceil-1$$
Hence, we have in particular that there are only $leftlceilfrac11-(x+varepsilon)rightrceil-1$ values of $n$ for which $u_nin[0,x+varepsilon[.$ Since all neighborhoods of $x$ have the form $[0,x+varepsilon[$ for some $varepsilonin,]0,1-x],$ then regardless of $x,$ every neighborhood of $x$ has only finitely-many $u_n$ inside it.
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
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Put simply, you can't, because it doesn't exist.
Take any $xin E.$ By definition of $E,$ we know that $0le x<1,$ and so by definition of $tau,$ the open neighborhoods of $x$ will be of the form $[0,y[$ for some $yin]x,1].$ However, taking $y=fracx+12,$ we see that only finitely-many $u_n$ will be in the set $[0,y[,$ so $u_n$ can't converge to $x.$
On the other hand, if we'd had the set $E=[0,1]$ and had $tau=bigl[0,x[::xin[0,1]bigrcupE,$ then we would find that $1$ is the limit of $u_n.$
Or, with the original $E$ and $tau,$ if we'd considered $v_n=frac1n+1,$ we would find that $v_n$ converges to every point of $E$!
Let me see if I can expand on my approach above.
First of all, note that for any $xin E$ and any $varepsilon,$ the following are equivalent: $$xin[0,x+varepsilon[;text and ;[0,x+varepsilon[,intau\x<x+varepsilonle 1\0<varepsilonle 1-x\varepsilonin,]0,1-x]$$
Next, note that for any $ninBbb N,$ any $xin E,$ and any $varepsilonin,]0,1-x],$ the following are equivalent: $$u_nin[0,x+varepsilon[\0le1-frac1n<x+varepsilon\frac1nle 1<frac1n+x+varepsilon\1le n<1+(x+varepsilon)n\n<1+(x+varepsilon)n\n-(x+varepsilon)n<1\nbigl(1-(x+varepsilon)bigr)<1\n<frac11-(x+varepsilon)\1le n<frac11-(x+varepsilon)\1le n<leftlceilfrac11-(x+varepsilon)rightrceil\1le nleleftlceilfrac11-(x+varepsilon)rightrceil-1$$
Hence, we have in particular that there are only $leftlceilfrac11-(x+varepsilon)rightrceil-1$ values of $n$ for which $u_nin[0,x+varepsilon[.$ Since all neighborhoods of $x$ have the form $[0,x+varepsilon[$ for some $varepsilonin,]0,1-x],$ then regardless of $x,$ every neighborhood of $x$ has only finitely-many $u_n$ inside it.
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
edited Mar 31 at 21:51
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
answered Mar 31 at 18:24
Cameron BuieCameron Buie
86.9k773161
86.9k773161
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
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– Poline Sandra
Mar 31 at 19:38
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sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
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– Poline Sandra
Mar 31 at 19:47
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Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
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– Cameron Buie
Mar 31 at 19:54
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I don't understand what you write
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– Poline Sandra
Mar 31 at 20:14
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Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
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– Cameron Buie
Mar 31 at 20:16
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show 5 more comments
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
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– Cameron Buie
Mar 31 at 19:54
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I don't understand what you write
$endgroup$
– Poline Sandra
Mar 31 at 20:14
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
I don't understand, in class we work with $varepsilon$: If I say let $lin E$ the open neighborhood is $[0,l+varepsilon[,varepsilon>0$, when $n>frac11-(l+varepsilon)$ the sequence is in $[0,l+varepsilon[$ how to prove that there is only a finit number? @Cameron Buie
$endgroup$
– Poline Sandra
Mar 31 at 19:38
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
sorry, $n<frac11-(l-varepsilon)$ with $l-varepsilonleq 1$ how to say that it is a finite number?
$endgroup$
– Poline Sandra
Mar 31 at 19:47
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
$endgroup$
– Cameron Buie
Mar 31 at 19:54
$begingroup$
Nice catch! Note that for $0<varepsilon<1-l,$ we will have that $frac11-(l+varepsilon)$ is a positive real number. But given any positive real number $x,$ there cannot be infinitely-many natural numbers less than $x.$ If there were, then $x$ would be greater than all of the natural numbers, but there isn't any real number greater than all of the natural numbers. Thus, there are only finitely-many $n$ less than $frac11-(l+varepsilon).$
$endgroup$
– Cameron Buie
Mar 31 at 19:54
$begingroup$
I don't understand what you write
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– Poline Sandra
Mar 31 at 20:14
$begingroup$
I don't understand what you write
$endgroup$
– Poline Sandra
Mar 31 at 20:14
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
$begingroup$
Well, what are some facts you know about the natural numbers? If you can give me something to go on, I may be able to clarify what I've written.
$endgroup$
– Cameron Buie
Mar 31 at 20:16
|
show 5 more comments
