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Is this set closed or open in $mathbbR$?
Open/closed sets in $[ 0, 1 ]$Is it false that the complement of an open set is closed?Proof of an open set or closed setTo determine if set is open or closedShowing this set is closed in $mathbbC$Question Involving Open/Closed SetsCan a set be both open and closed in vector space $mathbbR^n$?Both open and closed set.Is the set open? Closed?Is below set is open In $mathbbR^2$
$begingroup$
Determine if $S = x in mathbbR mid x^2 > x$ is closed or open
in $mathbbR$.
An answer key that I have says the set is closed. However, there is a theorem in this book that I have which says the following:
Let $f : mathbbR^n rightarrow mathbbR$ be continuous, and let $c in mathbbR$. Then the set $u in mathbbR^n mid f(u) > c$ is open in $mathbbR^n$.
So, why can't we just rewrite $S$ as $x in mathbbR mid x^2 - x > 0$. Then, shouldn't it be open? -- the function $f(x) = x^2 - x$ is continuous, and we can just take $c = 0$. What's wrong with my reasoning?
real-analysis general-topology
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Determine if $S = x in mathbbR mid x^2 > x$ is closed or open
in $mathbbR$.
An answer key that I have says the set is closed. However, there is a theorem in this book that I have which says the following:
Let $f : mathbbR^n rightarrow mathbbR$ be continuous, and let $c in mathbbR$. Then the set $u in mathbbR^n mid f(u) > c$ is open in $mathbbR^n$.
So, why can't we just rewrite $S$ as $x in mathbbR mid x^2 - x > 0$. Then, shouldn't it be open? -- the function $f(x) = x^2 - x$ is continuous, and we can just take $c = 0$. What's wrong with my reasoning?
real-analysis general-topology
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago
add a comment |
$begingroup$
Determine if $S = x in mathbbR mid x^2 > x$ is closed or open
in $mathbbR$.
An answer key that I have says the set is closed. However, there is a theorem in this book that I have which says the following:
Let $f : mathbbR^n rightarrow mathbbR$ be continuous, and let $c in mathbbR$. Then the set $u in mathbbR^n mid f(u) > c$ is open in $mathbbR^n$.
So, why can't we just rewrite $S$ as $x in mathbbR mid x^2 - x > 0$. Then, shouldn't it be open? -- the function $f(x) = x^2 - x$ is continuous, and we can just take $c = 0$. What's wrong with my reasoning?
real-analysis general-topology
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Determine if $S = x in mathbbR mid x^2 > x$ is closed or open
in $mathbbR$.
An answer key that I have says the set is closed. However, there is a theorem in this book that I have which says the following:
Let $f : mathbbR^n rightarrow mathbbR$ be continuous, and let $c in mathbbR$. Then the set $u in mathbbR^n mid f(u) > c$ is open in $mathbbR^n$.
So, why can't we just rewrite $S$ as $x in mathbbR mid x^2 - x > 0$. Then, shouldn't it be open? -- the function $f(x) = x^2 - x$ is continuous, and we can just take $c = 0$. What's wrong with my reasoning?
real-analysis general-topology
real-analysis general-topology
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
stackofhay42stackofhay42
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
stackofhay42stackofhay42
1823
asked 2 days ago
stackofhay42stackofhay42
1823
1823
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
stackofhay42 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago
add a comment |
$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago
$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago
$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Nothing is wrong with your reasoning. There's a mistake in the answer key.
answered 2 days ago
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Your reasoning is apt. Additionally your set is
$$S = (-infty,0)cup (1,infty).$$ Clearly the union of two open sets.
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
$endgroup$
add a comment |
$begingroup$
Nothing is wrong with your reasoning. The set $S$ is open.
In principle, it could be closed as well, but the only open and closed subsets of $Bbb R$ are $emptyset$ and $Bbb R$ (and $S$ is neither of these).
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
add a comment |
$begingroup$
If $x>1$ then $x^2>x.$ So take a sequence $a_n$ approaching $1$ from above. Then $a_n in S,$ but its limit is not. So set not closed.
answered 2 days ago
coffeemathcoffeemath
2,8871415
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. Observe that $x=1$ is a limit point not contained in your set. So it is definitely not closed.
answered 2 days ago
GReyesGReyes
1,93515
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nothing is wrong with your reasoning. There's a mistake in the answer key.
answered 2 days ago
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Nothing is wrong with your reasoning. There's a mistake in the answer key.
answered 2 days ago
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Nothing is wrong with your reasoning. There's a mistake in the answer key.
answered 2 days ago
jmerryjmerry
13.6k1629
$endgroup$
Nothing is wrong with your reasoning. There's a mistake in the answer key.
answered 2 days ago
jmerryjmerry
13.6k1629
answered 2 days ago
jmerryjmerry
13.6k1629
answered 2 days ago
jmerryjmerry
13.6k1629
answered 2 days ago
jmerryjmerry
13.6k1629
13.6k1629
add a comment |
add a comment |
$begingroup$
Your reasoning is apt. Additionally your set is
$$S = (-infty,0)cup (1,infty).$$ Clearly the union of two open sets.
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
$endgroup$
add a comment |
$begingroup$
Your reasoning is apt. Additionally your set is
$$S = (-infty,0)cup (1,infty).$$ Clearly the union of two open sets.
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
$endgroup$
add a comment |
$begingroup$
Your reasoning is apt. Additionally your set is
$$S = (-infty,0)cup (1,infty).$$ Clearly the union of two open sets.
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
$endgroup$
Your reasoning is apt. Additionally your set is
$$S = (-infty,0)cup (1,infty).$$ Clearly the union of two open sets.
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
answered 2 days ago
Gautam ShenoyGautam Shenoy
7,31411745
7,31411745
add a comment |
add a comment |
$begingroup$
Nothing is wrong with your reasoning. The set $S$ is open.
In principle, it could be closed as well, but the only open and closed subsets of $Bbb R$ are $emptyset$ and $Bbb R$ (and $S$ is neither of these).
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
add a comment |
$begingroup$
Nothing is wrong with your reasoning. The set $S$ is open.
In principle, it could be closed as well, but the only open and closed subsets of $Bbb R$ are $emptyset$ and $Bbb R$ (and $S$ is neither of these).
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
add a comment |
$begingroup$
Nothing is wrong with your reasoning. The set $S$ is open.
In principle, it could be closed as well, but the only open and closed subsets of $Bbb R$ are $emptyset$ and $Bbb R$ (and $S$ is neither of these).
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
Nothing is wrong with your reasoning. The set $S$ is open.
In principle, it could be closed as well, but the only open and closed subsets of $Bbb R$ are $emptyset$ and $Bbb R$ (and $S$ is neither of these).
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
add a comment |
add a comment |
$begingroup$
If $x>1$ then $x^2>x.$ So take a sequence $a_n$ approaching $1$ from above. Then $a_n in S,$ but its limit is not. So set not closed.
answered 2 days ago
coffeemathcoffeemath
2,8871415
$endgroup$
add a comment |
$begingroup$
If $x>1$ then $x^2>x.$ So take a sequence $a_n$ approaching $1$ from above. Then $a_n in S,$ but its limit is not. So set not closed.
answered 2 days ago
coffeemathcoffeemath
2,8871415
$endgroup$
add a comment |
$begingroup$
If $x>1$ then $x^2>x.$ So take a sequence $a_n$ approaching $1$ from above. Then $a_n in S,$ but its limit is not. So set not closed.
answered 2 days ago
coffeemathcoffeemath
2,8871415
$endgroup$
If $x>1$ then $x^2>x.$ So take a sequence $a_n$ approaching $1$ from above. Then $a_n in S,$ but its limit is not. So set not closed.
answered 2 days ago
coffeemathcoffeemath
2,8871415
answered 2 days ago
coffeemathcoffeemath
2,8871415
answered 2 days ago
coffeemathcoffeemath
2,8871415
answered 2 days ago
coffeemathcoffeemath
2,8871415
2,8871415
add a comment |
add a comment |
$begingroup$
Your reasoning is correct. Observe that $x=1$ is a limit point not contained in your set. So it is definitely not closed.
answered 2 days ago
GReyesGReyes
1,93515
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. Observe that $x=1$ is a limit point not contained in your set. So it is definitely not closed.
answered 2 days ago
GReyesGReyes
1,93515
$endgroup$
add a comment |
$begingroup$
Your reasoning is correct. Observe that $x=1$ is a limit point not contained in your set. So it is definitely not closed.
answered 2 days ago
GReyesGReyes
1,93515
$endgroup$
Your reasoning is correct. Observe that $x=1$ is a limit point not contained in your set. So it is definitely not closed.
answered 2 days ago
GReyesGReyes
1,93515
answered 2 days ago
GReyesGReyes
1,93515
answered 2 days ago
GReyesGReyes
1,93515
answered 2 days ago
GReyesGReyes
1,93515
1,93515
add a comment |
add a comment |
stackofhay42 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You are correct and the answer key is wrong
$endgroup$
– Math_QED
2 days ago