Is every $xin (0,1) setminus S$ is a limit point of $S$?
up vote
0
down vote
favorite
let $S$ be the set of all rational number in $(0,1)$
.
then which of the following statement is True
$1)$$ S$ is not a closed subset of $ mathbb{R}$
$2$) $S$ is closed subset of $mathbb{R}$
$3)$ S is an open subset of $mathbb{R}$
$4)$ Every $xin (0,1) setminus S$ is a limit point of $S$
My attempt : option $1$ and option $3)$ will correct because here $[0,1]$ is a limit point of S
im confusing about option $4$
Any hints/solution will be appreciated ?
real-analysis
asked Nov 16 at 13:23
jasmine
1,326416
add a comment |
up vote
0
down vote
favorite
let $S$ be the set of all rational number in $(0,1)$
.
then which of the following statement is True
$1)$$ S$ is not a closed subset of $ mathbb{R}$
$2$) $S$ is closed subset of $mathbb{R}$
$3)$ S is an open subset of $mathbb{R}$
$4)$ Every $xin (0,1) setminus S$ is a limit point of $S$
My attempt : option $1$ and option $3)$ will correct because here $[0,1]$ is a limit point of S
im confusing about option $4$
Any hints/solution will be appreciated ?
real-analysis
asked Nov 16 at 13:23
jasmine
1,326416
1
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
let $S$ be the set of all rational number in $(0,1)$
.
then which of the following statement is True
$1)$$ S$ is not a closed subset of $ mathbb{R}$
$2$) $S$ is closed subset of $mathbb{R}$
$3)$ S is an open subset of $mathbb{R}$
$4)$ Every $xin (0,1) setminus S$ is a limit point of $S$
My attempt : option $1$ and option $3)$ will correct because here $[0,1]$ is a limit point of S
im confusing about option $4$
Any hints/solution will be appreciated ?
real-analysis
asked Nov 16 at 13:23
jasmine
1,326416
let $S$ be the set of all rational number in $(0,1)$
.
then which of the following statement is True
$1)$$ S$ is not a closed subset of $ mathbb{R}$
$2$) $S$ is closed subset of $mathbb{R}$
$3)$ S is an open subset of $mathbb{R}$
$4)$ Every $xin (0,1) setminus S$ is a limit point of $S$
My attempt : option $1$ and option $3)$ will correct because here $[0,1]$ is a limit point of S
im confusing about option $4$
Any hints/solution will be appreciated ?
real-analysis
real-analysis
asked Nov 16 at 13:23
jasmine
1,326416
asked Nov 16 at 13:23
jasmine
1,326416
asked Nov 16 at 13:23
jasmine
1,326416
asked Nov 16 at 13:23
jasmine
1,326416
asked Nov 16 at 13:23
jasmine
1,326416
1,326416
1
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26
add a comment |
1
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26
1
1
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
1) True, as the complement does not contain an open ball. If it did, let the radius of such a ball be $epsilon$ and consider rational numbers of denominator $>1/epsilon$; at least one of them must be in there
2) False
3) False; contains no open ball
4) True, e.g. $alpha$ irrational and consider $alpha+1/n$, $ninmathbb N$.
answered Nov 16 at 13:43
Richard Martin
1,5468
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
1) True, as the complement does not contain an open ball. If it did, let the radius of such a ball be $epsilon$ and consider rational numbers of denominator $>1/epsilon$; at least one of them must be in there
2) False
3) False; contains no open ball
4) True, e.g. $alpha$ irrational and consider $alpha+1/n$, $ninmathbb N$.
answered Nov 16 at 13:43
Richard Martin
1,5468
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
add a comment |
up vote
1
down vote
accepted
1) True, as the complement does not contain an open ball. If it did, let the radius of such a ball be $epsilon$ and consider rational numbers of denominator $>1/epsilon$; at least one of them must be in there
2) False
3) False; contains no open ball
4) True, e.g. $alpha$ irrational and consider $alpha+1/n$, $ninmathbb N$.
answered Nov 16 at 13:43
Richard Martin
1,5468
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
1) True, as the complement does not contain an open ball. If it did, let the radius of such a ball be $epsilon$ and consider rational numbers of denominator $>1/epsilon$; at least one of them must be in there
2) False
3) False; contains no open ball
4) True, e.g. $alpha$ irrational and consider $alpha+1/n$, $ninmathbb N$.
answered Nov 16 at 13:43
Richard Martin
1,5468
1) True, as the complement does not contain an open ball. If it did, let the radius of such a ball be $epsilon$ and consider rational numbers of denominator $>1/epsilon$; at least one of them must be in there
2) False
3) False; contains no open ball
4) True, e.g. $alpha$ irrational and consider $alpha+1/n$, $ninmathbb N$.
answered Nov 16 at 13:43
Richard Martin
1,5468
edited Nov 16 at 14:03
answered Nov 16 at 13:43
Richard Martin
1,5468
answered Nov 16 at 13:43
Richard Martin
1,5468
answered Nov 16 at 13:43
Richard Martin
1,5468
1,5468
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
add a comment |
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
can u little bit elaborate how it does not contain open ball ??
– jasmine
Nov 16 at 13:57
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001130%2fis-every-x-in-0-1-setminus-s-is-a-limit-point-of-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Do you know/are you allowed to cite the fact that any non-empty, open subinterval of $mathbb{R}$ contains a rational number?
– bangs
Nov 16 at 13:26