Checking Continuity of functions
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Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions
continuity
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Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions
continuity
3
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions
continuity
Consider a function $sqrt x-1$+$sqrt1-x$.From here we can see that domain of the function is just 1 and range is 0.Still the function is continious at x=1 even though RHL and LHL limit doesn't exist. Can you guys tell me how this is possible. What is the criterion for checking continuity of functions
continuity
edited Aug 19 at 9:10
asked Aug 19 at 9:09
user501655
3
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10
add a comment |Â
3
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10
3
3
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10
add a comment |Â
3 Answers
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1
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If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if
(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$
or
(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).
In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.
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What is your definition for $l=limlimits_xto 1f(x)$ if it exists?
If this is it:
$$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$
then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.
add a comment |Â
up vote
0
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Continuity is a topological concept.
If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.
So that is the criterion that should be handled here.
Observe that limits do not play a part in it.
In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).
Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.
It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.
To get more view on this you could take a look at this answer.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if
(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$
or
(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).
In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.
add a comment |Â
up vote
1
down vote
If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if
(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$
or
(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).
In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if
(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$
or
(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).
In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.
If you're using the definition of $lim_x to a f(x)$ where the value $f(a)$ (if defined) should not be taken into account, then $f$ satisfies the $varepsilon$/$delta$ definition of being continuous at the point $a in D_f$ if and only if
(1) $a$ is an accumulation point of the domain $D_f$, and $f(x) to f(a)$ as $x to a$
or
(2) $a$ is an isolated point of $D_f$ (in which case it's not meaningful to talk about $lim_x to a f(x)$).
In your example, statement (2) holds, so $f$ is continuous at the point $a=1$.
answered Aug 19 at 9:55
Hans Lundmark
33.2k564109
33.2k564109
add a comment |Â
add a comment |Â
up vote
0
down vote
What is your definition for $l=limlimits_xto 1f(x)$ if it exists?
If this is it:
$$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$
then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.
add a comment |Â
up vote
0
down vote
What is your definition for $l=limlimits_xto 1f(x)$ if it exists?
If this is it:
$$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$
then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What is your definition for $l=limlimits_xto 1f(x)$ if it exists?
If this is it:
$$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$
then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.
What is your definition for $l=limlimits_xto 1f(x)$ if it exists?
If this is it:
$$forallvarepsilon>0,existsdelta>0,forall xin D_f,|x-1|<deltaimplies|f(x)-f(1)|<varepsilon$$
then take ANY $varepsilon>0$ and ANY $delta>0$. $D_f=1$ so $|x-1|=0$ for any $xin D_f$ and thus $f(x)=f(1)$.
answered Aug 19 at 9:29
Scientifica
4,90121231
4,90121231
add a comment |Â
add a comment |Â
up vote
0
down vote
Continuity is a topological concept.
If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.
So that is the criterion that should be handled here.
Observe that limits do not play a part in it.
In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).
Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.
It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.
To get more view on this you could take a look at this answer.
add a comment |Â
up vote
0
down vote
Continuity is a topological concept.
If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.
So that is the criterion that should be handled here.
Observe that limits do not play a part in it.
In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).
Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.
It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.
To get more view on this you could take a look at this answer.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Continuity is a topological concept.
If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.
So that is the criterion that should be handled here.
Observe that limits do not play a part in it.
In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).
Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.
It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.
To get more view on this you could take a look at this answer.
Continuity is a topological concept.
If $langle X,tau_Xrangle$ and $langle Y,tau_Yrangle$ are topological spaces and $f:Xto Y$ is a function then $f$ is by definition continuous if all preimages of elements $tau_Ysubseteqwp(Y)$ under $f$ are elements of $tau_Xsubseteqwp(X)$.
So that is the criterion that should be handled here.
Observe that limits do not play a part in it.
In this situation $X=1$ and then automatically we must have $tau_X=wp(X)$ (there are no other topologies on $1$).
Then it is immediate that any function that has $1$ as domain is continuous, because preimages under that function will of course be elements of $wp(1)=varnothing,1$.
It might well be that you are not familiar with topology and that - in spite of that - still are familiar somehow with continuity. The main cause of this is that continuity already shows up in metric spaces, and these spaces can be handled without any reference to topology. Also they carry the possibility of a definition of continuity. But actually metric spaces induce a topology and the definition of continuity in this topological space coincides with the one in the metric space.
To get more view on this you could take a look at this answer.
answered Aug 19 at 9:35
drhab
87.7k541119
87.7k541119
add a comment |Â
add a comment |Â
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3
If $X$ and $Y$ are topological spaces, and $X$ has just one point, then all maps $Xto Y$ are continuous.
â Lord Shark the Unknown
Aug 19 at 9:10