An example of a sequence in $mathbb{R}$ whose set of accumulation point is $mathbb{N}$
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I'm trying to find an example of a sequence in $mathbb{R}$ whose set of accumulation point is $mathbb{N}$, or whose set of accumulation point is $mathbb{R}$.
I would like to show my progress but I'm stuck.
real-analysis
asked Nov 18 at 11:20
user15269
1608
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up vote
0
down vote
favorite
I'm trying to find an example of a sequence in $mathbb{R}$ whose set of accumulation point is $mathbb{N}$, or whose set of accumulation point is $mathbb{R}$.
I would like to show my progress but I'm stuck.
real-analysis
asked Nov 18 at 11:20
user15269
1608
For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to find an example of a sequence in $mathbb{R}$ whose set of accumulation point is $mathbb{N}$, or whose set of accumulation point is $mathbb{R}$.
I would like to show my progress but I'm stuck.
real-analysis
asked Nov 18 at 11:20
user15269
1608
I'm trying to find an example of a sequence in $mathbb{R}$ whose set of accumulation point is $mathbb{N}$, or whose set of accumulation point is $mathbb{R}$.
I would like to show my progress but I'm stuck.
real-analysis
real-analysis
asked Nov 18 at 11:20
user15269
1608
asked Nov 18 at 11:20
user15269
1608
asked Nov 18 at 11:20
user15269
1608
asked Nov 18 at 11:20
user15269
1608
asked Nov 18 at 11:20
user15269
1608
1608
For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22
add a comment |
For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22
For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22
For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22
add a comment |
1 Answer
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For $mathbb{R}$: take a bijection $f:mathbb{N}tomathbb{Q}$. Take the sequence whose $n$th term is $f(n)$. This works because we obtain every rational number, and every real number has a sequence of distinct rational numbers converging to it.
For $mathbb{N}$: Take a bijection $g: mathbb{N}tomathbb{N^2}$. Take the sequence whose $n$th term is the first coordinate of $g(n)$. This works because we obtain every natural number infinitely many times.
answered Nov 18 at 11:24
user3482749
1,916411
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For $mathbb{R}$: take a bijection $f:mathbb{N}tomathbb{Q}$. Take the sequence whose $n$th term is $f(n)$. This works because we obtain every rational number, and every real number has a sequence of distinct rational numbers converging to it.
For $mathbb{N}$: Take a bijection $g: mathbb{N}tomathbb{N^2}$. Take the sequence whose $n$th term is the first coordinate of $g(n)$. This works because we obtain every natural number infinitely many times.
answered Nov 18 at 11:24
user3482749
1,916411
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
add a comment |
up vote
0
down vote
For $mathbb{R}$: take a bijection $f:mathbb{N}tomathbb{Q}$. Take the sequence whose $n$th term is $f(n)$. This works because we obtain every rational number, and every real number has a sequence of distinct rational numbers converging to it.
For $mathbb{N}$: Take a bijection $g: mathbb{N}tomathbb{N^2}$. Take the sequence whose $n$th term is the first coordinate of $g(n)$. This works because we obtain every natural number infinitely many times.
answered Nov 18 at 11:24
user3482749
1,916411
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
add a comment |
up vote
0
down vote
up vote
0
down vote
For $mathbb{R}$: take a bijection $f:mathbb{N}tomathbb{Q}$. Take the sequence whose $n$th term is $f(n)$. This works because we obtain every rational number, and every real number has a sequence of distinct rational numbers converging to it.
For $mathbb{N}$: Take a bijection $g: mathbb{N}tomathbb{N^2}$. Take the sequence whose $n$th term is the first coordinate of $g(n)$. This works because we obtain every natural number infinitely many times.
answered Nov 18 at 11:24
user3482749
1,916411
For $mathbb{R}$: take a bijection $f:mathbb{N}tomathbb{Q}$. Take the sequence whose $n$th term is $f(n)$. This works because we obtain every rational number, and every real number has a sequence of distinct rational numbers converging to it.
For $mathbb{N}$: Take a bijection $g: mathbb{N}tomathbb{N^2}$. Take the sequence whose $n$th term is the first coordinate of $g(n)$. This works because we obtain every natural number infinitely many times.
answered Nov 18 at 11:24
user3482749
1,916411
answered Nov 18 at 11:24
user3482749
1,916411
answered Nov 18 at 11:24
user3482749
1,916411
answered Nov 18 at 11:24
user3482749
1,916411
1,916411
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
add a comment |
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
An explicit example: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, …
– Patrick Stevens
Nov 18 at 11:26
add a comment |
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For the second one you can take $;Bbb Q;$ ...
– DonAntonio
Nov 18 at 11:22