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Counterexample to: “If a function is continuous in a point $x_0$ then it is defined in a neighborhood of...
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I'm looking for a counterexample to the statement:
If a function is continuous in a point $x_0$ then it is defined in a
neighborhood of that point.
If I take
$f: mathbb{R} setminus mathbb{Q} rightarrow mathbb{R}$ defined as $f(x):=x$
Then it is continuous in any point $x_0 in mathbb{R}setminus mathbb{Q}$ since $lim_{xto x_0} f(x)=x_0 forall x_0 in mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 in mathbb{R}$.
Is this counterexample correct?
Edit
Definition of neighborhood of $x_0$: Let $x_0 in mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $mathbb{R}$ $(x_0-r,x_0+r)$.
Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x to x_0} f(x)=l iff forall epsilon >0 exists delta >0 mid forall xin domf wedge 0<mid x-x_0 mid <delta implies mid f(x)-lmid <epsilon$
Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $forall epsilon >0 exists delta>0 mid forall x in dom(f) wedge mid x-x_0 mid < delta implies mid f(x)-f(x_0)mid <epsilon$
calculus continuity
asked Jan 6 '16 at 16:56
Gianolepo
731918
|
show 6 more comments
up vote
0
down vote
favorite
I'm looking for a counterexample to the statement:
If a function is continuous in a point $x_0$ then it is defined in a
neighborhood of that point.
If I take
$f: mathbb{R} setminus mathbb{Q} rightarrow mathbb{R}$ defined as $f(x):=x$
Then it is continuous in any point $x_0 in mathbb{R}setminus mathbb{Q}$ since $lim_{xto x_0} f(x)=x_0 forall x_0 in mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 in mathbb{R}$.
Is this counterexample correct?
Edit
Definition of neighborhood of $x_0$: Let $x_0 in mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $mathbb{R}$ $(x_0-r,x_0+r)$.
Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x to x_0} f(x)=l iff forall epsilon >0 exists delta >0 mid forall xin domf wedge 0<mid x-x_0 mid <delta implies mid f(x)-lmid <epsilon$
Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $forall epsilon >0 exists delta>0 mid forall x in dom(f) wedge mid x-x_0 mid < delta implies mid f(x)-f(x_0)mid <epsilon$
calculus continuity
asked Jan 6 '16 at 16:56
Gianolepo
731918
What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
2
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm looking for a counterexample to the statement:
If a function is continuous in a point $x_0$ then it is defined in a
neighborhood of that point.
If I take
$f: mathbb{R} setminus mathbb{Q} rightarrow mathbb{R}$ defined as $f(x):=x$
Then it is continuous in any point $x_0 in mathbb{R}setminus mathbb{Q}$ since $lim_{xto x_0} f(x)=x_0 forall x_0 in mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 in mathbb{R}$.
Is this counterexample correct?
Edit
Definition of neighborhood of $x_0$: Let $x_0 in mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $mathbb{R}$ $(x_0-r,x_0+r)$.
Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x to x_0} f(x)=l iff forall epsilon >0 exists delta >0 mid forall xin domf wedge 0<mid x-x_0 mid <delta implies mid f(x)-lmid <epsilon$
Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $forall epsilon >0 exists delta>0 mid forall x in dom(f) wedge mid x-x_0 mid < delta implies mid f(x)-f(x_0)mid <epsilon$
calculus continuity
asked Jan 6 '16 at 16:56
Gianolepo
731918
I'm looking for a counterexample to the statement:
If a function is continuous in a point $x_0$ then it is defined in a
neighborhood of that point.
If I take
$f: mathbb{R} setminus mathbb{Q} rightarrow mathbb{R}$ defined as $f(x):=x$
Then it is continuous in any point $x_0 in mathbb{R}setminus mathbb{Q}$ since $lim_{xto x_0} f(x)=x_0 forall x_0 in mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 in mathbb{R}$.
Is this counterexample correct?
Edit
Definition of neighborhood of $x_0$: Let $x_0 in mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $mathbb{R}$ $(x_0-r,x_0+r)$.
Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x to x_0} f(x)=l iff forall epsilon >0 exists delta >0 mid forall xin domf wedge 0<mid x-x_0 mid <delta implies mid f(x)-lmid <epsilon$
Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $forall epsilon >0 exists delta>0 mid forall x in dom(f) wedge mid x-x_0 mid < delta implies mid f(x)-f(x_0)mid <epsilon$
calculus continuity
calculus continuity
asked Jan 6 '16 at 16:56
Gianolepo
731918
asked Jan 6 '16 at 16:56
Gianolepo
731918
edited Jan 6 '16 at 17:44
asked Jan 6 '16 at 16:56
Gianolepo
731918
asked Jan 6 '16 at 16:56
Gianolepo
731918
asked Jan 6 '16 at 16:56
Gianolepo
731918
731918
What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
2
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25
|
show 6 more comments
What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
2
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25
What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
2
2
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25
|
show 6 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.
Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that
$$d_Y(f(x),f(y)) < varepsilon$$
whenever
$$yin X, ~ d_X(x,y) < delta$$
Note that this definition is equivalent to yours when $x$ is a limit point of the domain.
answered Jan 6 '16 at 17:12
B. Freitas
491212
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
add a comment |
up vote
0
down vote
If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.
answered yesterday
Richard Martin
1,3438
add a comment |
up vote
-2
down vote
No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.
$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$
(where $V_a(b)$ is the open $a$-neighbourhood of $b$)
For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.
answered Jan 6 '16 at 17:15
user1892304
1,402817
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.
Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that
$$d_Y(f(x),f(y)) < varepsilon$$
whenever
$$yin X, ~ d_X(x,y) < delta$$
Note that this definition is equivalent to yours when $x$ is a limit point of the domain.
answered Jan 6 '16 at 17:12
B. Freitas
491212
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
add a comment |
up vote
1
down vote
Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.
Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that
$$d_Y(f(x),f(y)) < varepsilon$$
whenever
$$yin X, ~ d_X(x,y) < delta$$
Note that this definition is equivalent to yours when $x$ is a limit point of the domain.
answered Jan 6 '16 at 17:12
B. Freitas
491212
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
add a comment |
up vote
1
down vote
up vote
1
down vote
Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.
Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that
$$d_Y(f(x),f(y)) < varepsilon$$
whenever
$$yin X, ~ d_X(x,y) < delta$$
Note that this definition is equivalent to yours when $x$ is a limit point of the domain.
answered Jan 6 '16 at 17:12
B. Freitas
491212
Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.
Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that
$$d_Y(f(x),f(y)) < varepsilon$$
whenever
$$yin X, ~ d_X(x,y) < delta$$
Note that this definition is equivalent to yours when $x$ is a limit point of the domain.
answered Jan 6 '16 at 17:12
B. Freitas
491212
answered Jan 6 '16 at 17:12
B. Freitas
491212
answered Jan 6 '16 at 17:12
B. Freitas
491212
answered Jan 6 '16 at 17:12
B. Freitas
491212
491212
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
add a comment |
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
Well, your definition requires $f$ to be defined on all of $X$. That's not the case for our OP, who is not talking about neighborhoods in $X$ but in some space containing $X$.
– Thomas Andrews
Jan 6 '16 at 17:29
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
That's precisely the point. The neighb. in this bigger space have nothing to do with the continuity of $f$. Also, $X$ is just the domain of $f$ so I don't understand what you mean by his definition not requiring $f$ to be defined in $X$.
– B. Freitas
Jan 6 '16 at 17:56
add a comment |
up vote
0
down vote
If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.
answered yesterday
Richard Martin
1,3438
add a comment |
up vote
0
down vote
If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.
answered yesterday
Richard Martin
1,3438
add a comment |
up vote
0
down vote
up vote
0
down vote
If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.
answered yesterday
Richard Martin
1,3438
If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
1,3438
add a comment |
add a comment |
up vote
-2
down vote
No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.
$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$
(where $V_a(b)$ is the open $a$-neighbourhood of $b$)
For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.
answered Jan 6 '16 at 17:15
user1892304
1,402817
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
add a comment |
up vote
-2
down vote
No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.
$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$
(where $V_a(b)$ is the open $a$-neighbourhood of $b$)
For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.
answered Jan 6 '16 at 17:15
user1892304
1,402817
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
add a comment |
up vote
-2
down vote
up vote
-2
down vote
No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.
$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$
(where $V_a(b)$ is the open $a$-neighbourhood of $b$)
For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.
answered Jan 6 '16 at 17:15
user1892304
1,402817
No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.
$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$
(where $V_a(b)$ is the open $a$-neighbourhood of $b$)
For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.
answered Jan 6 '16 at 17:15
user1892304
1,402817
answered Jan 6 '16 at 17:15
user1892304
1,402817
answered Jan 6 '16 at 17:15
user1892304
1,402817
answered Jan 6 '16 at 17:15
user1892304
1,402817
1,402817
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
add a comment |
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
Since OP is talking about limit points I assume his definition of limit is not so naive.
– B. Freitas
Jan 6 '16 at 17:19
add a comment |
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What is your definition of neighborhood?
– B. Freitas
Jan 6 '16 at 16:58
2
The word "neighborhood" is relative to a set. So it is unclear what you are really asking. For example, the neighborhoods of $0$ in $mathbb Q$ are different from the neighborhoods of $0$ in $mathbb R$. So it is definitely possible to define a function on $mathbb Q$ that is continuous at $0$, but isn't defined on the irrationals. But it is still defined on a neighborhood of $0$ in $mathbb Q$.
– Thomas Andrews
Jan 6 '16 at 17:00
Note that it is always possible to have "$xto x_0$" without requiring there be an interval of values through which $x$ travels (I'm assuming you're thinking about the real line). You can approach a point through discrete points, or even through a finite number (including only the point itself) of points.
– MPW
Jan 6 '16 at 17:04
"Defininition of continuity that uses limits" is not actually a useful definition. What does the definition mean, when there are lots of points where the function isn't defined? What does $lim_{xto a} f(x)$ mean then?
– Thomas Andrews
Jan 6 '16 at 17:05
@ThomasAndrews I added the "epsilon-delta" definition, but, as far as I know, that is equivalent to the condition $lim_{xto x_0} f(x)=f(x_0)$ if $x_0$ is a limit point in the domain of $f$ (and that is the case). The definition of limit does not require that the function is defined in every point of a neighborhood of $x_0$
– Gianolepo
Jan 6 '16 at 17:25