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$f$ is continuous in $mathbbR$ if and only if for every open set $G$, the set $f^-1(G)=x:f(x)in G $ is open. [duplicate]
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This question already has an answer here:
Prove $epsilon$-$delta$ definition of continuity implies the open set definition for real function
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Suppose $f(x)$ defined in $mathbbR$, then $f$ is continuous in $mathbbR$ if and only if for every open set $G$, the set $f^-1(G)=x:f(x)in G $ is open.
I have done necessity.
If $f$ is continuous and G is open. Then for $x_0in f^-1(G)$,$f(x_0)in G$,and $f$ is continuous hence $exists U(x_0,delta)$,such that for $xin U(x_0,delta)$,$f(x)in G$. Thus $ U(x_0,delta)subset f^-1(G)$ which makes $f^-1(G)$ is an open set.
And now I have no ideal of the sufficiency.
real-analysis general-topology
asked Jul 17 at 21:12
Jaqen Chou
1526
marked as duplicate by amWhy, max_zorn, Rhys Steele, Delta-u, Gibbs Jul 18 at 9:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
Prove $epsilon$-$delta$ definition of continuity implies the open set definition for real function
1 answer
Suppose $f(x)$ defined in $mathbbR$, then $f$ is continuous in $mathbbR$ if and only if for every open set $G$, the set $f^-1(G)=x:f(x)in G $ is open.
I have done necessity.
If $f$ is continuous and G is open. Then for $x_0in f^-1(G)$,$f(x_0)in G$,and $f$ is continuous hence $exists U(x_0,delta)$,such that for $xin U(x_0,delta)$,$f(x)in G$. Thus $ U(x_0,delta)subset f^-1(G)$ which makes $f^-1(G)$ is an open set.
And now I have no ideal of the sufficiency.
real-analysis general-topology
asked Jul 17 at 21:12
Jaqen Chou
1526
marked as duplicate by amWhy, max_zorn, Rhys Steele, Delta-u, Gibbs Jul 18 at 9:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
Prove $epsilon$-$delta$ definition of continuity implies the open set definition for real function
1 answer
Suppose $f(x)$ defined in $mathbbR$, then $f$ is continuous in $mathbbR$ if and only if for every open set $G$, the set $f^-1(G)=x:f(x)in G $ is open.
I have done necessity.
If $f$ is continuous and G is open. Then for $x_0in f^-1(G)$,$f(x_0)in G$,and $f$ is continuous hence $exists U(x_0,delta)$,such that for $xin U(x_0,delta)$,$f(x)in G$. Thus $ U(x_0,delta)subset f^-1(G)$ which makes $f^-1(G)$ is an open set.
And now I have no ideal of the sufficiency.
real-analysis general-topology
asked Jul 17 at 21:12
Jaqen Chou
1526
This question already has an answer here:
Prove $epsilon$-$delta$ definition of continuity implies the open set definition for real function
1 answer
Suppose $f(x)$ defined in $mathbbR$, then $f$ is continuous in $mathbbR$ if and only if for every open set $G$, the set $f^-1(G)=x:f(x)in G $ is open.
I have done necessity.
If $f$ is continuous and G is open. Then for $x_0in f^-1(G)$,$f(x_0)in G$,and $f$ is continuous hence $exists U(x_0,delta)$,such that for $xin U(x_0,delta)$,$f(x)in G$. Thus $ U(x_0,delta)subset f^-1(G)$ which makes $f^-1(G)$ is an open set.
And now I have no ideal of the sufficiency.
This question already has an answer here:
Prove $epsilon$-$delta$ definition of continuity implies the open set definition for real function
1 answer
real-analysis general-topology
asked Jul 17 at 21:12
Jaqen Chou
1526
asked Jul 17 at 21:12
Jaqen Chou
1526
asked Jul 17 at 21:12
Jaqen Chou
1526
asked Jul 17 at 21:12
Jaqen Chou
1526
1526
marked as duplicate by amWhy, max_zorn, Rhys Steele, Delta-u, Gibbs Jul 18 at 9:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy, max_zorn, Rhys Steele, Delta-u, Gibbs Jul 18 at 9:06
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38
add a comment |Â
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Hint: Fix $x in mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $epsilon > 0$ and consider the open set $(f(x) - epsilon, f(x) + epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
add a comment |Â
up vote
0
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If $x_0 in X$ and $varepsilon>0$ the set $G = B(f(x_0), varepsilon)$ is open.
So $f^-1[G]$ is open and $x_0 in f^-1[G]$ as $f(x_0) in G = B(f(x_0), varepsilon)$ clearly.
So there is some $delta>0$ such that $B(x_0, delta) subseteq f^-1[G]$ and it's easy to see that this $delta>0$ is as required for our starting $varepsilon$. So $f$ is continuous at $x_0$.
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: Fix $x in mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $epsilon > 0$ and consider the open set $(f(x) - epsilon, f(x) + epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
add a comment |Â
up vote
2
down vote
accepted
Hint: Fix $x in mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $epsilon > 0$ and consider the open set $(f(x) - epsilon, f(x) + epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: Fix $x in mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $epsilon > 0$ and consider the open set $(f(x) - epsilon, f(x) + epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
Hint: Fix $x in mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $epsilon > 0$ and consider the open set $(f(x) - epsilon, f(x) + epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
answered Jul 17 at 21:21
Stefan Mesken
13.6k32045
13.6k32045
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
add a comment |Â
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
the image of the open interval contains $x$ subset $(f(x)-epsilon,f(x)+epsilon)$.
– Jaqen Chou
Jul 17 at 21:36
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
@JaqenChou You need to show that the $epsilon$-$delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $delta$ here?
– Stefan Mesken
Jul 17 at 21:40
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
can we just say that there exists a open interval $B(x,delta)$ subset the preimage since the preimage is open and has the element x.
– Jaqen Chou
Jul 17 at 21:50
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
@JaqenChou Yep. And that $delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.)
– Stefan Mesken
Jul 17 at 21:52
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
good teacher of you~ Memeda
– Jaqen Chou
Jul 17 at 21:54
add a comment |Â
up vote
0
down vote
If $x_0 in X$ and $varepsilon>0$ the set $G = B(f(x_0), varepsilon)$ is open.
So $f^-1[G]$ is open and $x_0 in f^-1[G]$ as $f(x_0) in G = B(f(x_0), varepsilon)$ clearly.
So there is some $delta>0$ such that $B(x_0, delta) subseteq f^-1[G]$ and it's easy to see that this $delta>0$ is as required for our starting $varepsilon$. So $f$ is continuous at $x_0$.
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
add a comment |Â
up vote
0
down vote
If $x_0 in X$ and $varepsilon>0$ the set $G = B(f(x_0), varepsilon)$ is open.
So $f^-1[G]$ is open and $x_0 in f^-1[G]$ as $f(x_0) in G = B(f(x_0), varepsilon)$ clearly.
So there is some $delta>0$ such that $B(x_0, delta) subseteq f^-1[G]$ and it's easy to see that this $delta>0$ is as required for our starting $varepsilon$. So $f$ is continuous at $x_0$.
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $x_0 in X$ and $varepsilon>0$ the set $G = B(f(x_0), varepsilon)$ is open.
So $f^-1[G]$ is open and $x_0 in f^-1[G]$ as $f(x_0) in G = B(f(x_0), varepsilon)$ clearly.
So there is some $delta>0$ such that $B(x_0, delta) subseteq f^-1[G]$ and it's easy to see that this $delta>0$ is as required for our starting $varepsilon$. So $f$ is continuous at $x_0$.
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
If $x_0 in X$ and $varepsilon>0$ the set $G = B(f(x_0), varepsilon)$ is open.
So $f^-1[G]$ is open and $x_0 in f^-1[G]$ as $f(x_0) in G = B(f(x_0), varepsilon)$ clearly.
So there is some $delta>0$ such that $B(x_0, delta) subseteq f^-1[G]$ and it's easy to see that this $delta>0$ is as required for our starting $varepsilon$. So $f$ is continuous at $x_0$.
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
answered Jul 17 at 21:55
Henno Brandsma
91.6k342100
91.6k342100
add a comment |Â
add a comment |Â
To show f is continuous what are you allowed to use? only $epsilon - delta$?
– gd1035
Jul 17 at 21:18
@ gd1035 If possible, $epsilon-delta$ is good for me to understand.
– Jaqen Chou
Jul 17 at 21:38