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$f$ is continuous, if $f_n$ continuous and $f_nto f$ uniformly
Clash Royale CLAN TAG#URR8PPP
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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: Xto Y$ with $ninmathbbN$ and $f:Xto Y$ functions. $f_n$ is continuous for every $n$ and $f_nstackrelntoinftylongrightarrow f$ uniformly.
Then $f$ is continuous.
Here is my proof:
Let $xin X$ and $epsilon >0$ be arbitrary.
Since $f_nto f$ uniformly it exists $NinmathbbN$ such that $d_Y(f_N(x), f(x))<epsilon/3$ for every $xin X$.
As $f_N$ is continuous we have for $x_0in X$ and $delta > 0$, that $d_Y(f_N(x),f_N(x_0))<epsilon/3$ if $d_X(x,x_0)<delta$.
This gives us:
$$beginalignd_Y(f(x),f(x_0)&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\
&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\
&< epsilon/3+epsilon/3+epsilon/3=epsilonendalign$$
Thanks in advance for your correction.
proof-verification metric-spaces continuity
asked Jul 17 at 18:38
Cornman
2,51921128
add a comment |Â
up vote
4
down vote
favorite
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: Xto Y$ with $ninmathbbN$ and $f:Xto Y$ functions. $f_n$ is continuous for every $n$ and $f_nstackrelntoinftylongrightarrow f$ uniformly.
Then $f$ is continuous.
Here is my proof:
Let $xin X$ and $epsilon >0$ be arbitrary.
Since $f_nto f$ uniformly it exists $NinmathbbN$ such that $d_Y(f_N(x), f(x))<epsilon/3$ for every $xin X$.
As $f_N$ is continuous we have for $x_0in X$ and $delta > 0$, that $d_Y(f_N(x),f_N(x_0))<epsilon/3$ if $d_X(x,x_0)<delta$.
This gives us:
$$beginalignd_Y(f(x),f(x_0)&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\
&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\
&< epsilon/3+epsilon/3+epsilon/3=epsilonendalign$$
Thanks in advance for your correction.
proof-verification metric-spaces continuity
asked Jul 17 at 18:38
Cornman
2,51921128
1
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
1
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: Xto Y$ with $ninmathbbN$ and $f:Xto Y$ functions. $f_n$ is continuous for every $n$ and $f_nstackrelntoinftylongrightarrow f$ uniformly.
Then $f$ is continuous.
Here is my proof:
Let $xin X$ and $epsilon >0$ be arbitrary.
Since $f_nto f$ uniformly it exists $NinmathbbN$ such that $d_Y(f_N(x), f(x))<epsilon/3$ for every $xin X$.
As $f_N$ is continuous we have for $x_0in X$ and $delta > 0$, that $d_Y(f_N(x),f_N(x_0))<epsilon/3$ if $d_X(x,x_0)<delta$.
This gives us:
$$beginalignd_Y(f(x),f(x_0)&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\
&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\
&< epsilon/3+epsilon/3+epsilon/3=epsilonendalign$$
Thanks in advance for your correction.
proof-verification metric-spaces continuity
asked Jul 17 at 18:38
Cornman
2,51921128
Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: Xto Y$ with $ninmathbbN$ and $f:Xto Y$ functions. $f_n$ is continuous for every $n$ and $f_nstackrelntoinftylongrightarrow f$ uniformly.
Then $f$ is continuous.
Here is my proof:
Let $xin X$ and $epsilon >0$ be arbitrary.
Since $f_nto f$ uniformly it exists $NinmathbbN$ such that $d_Y(f_N(x), f(x))<epsilon/3$ for every $xin X$.
As $f_N$ is continuous we have for $x_0in X$ and $delta > 0$, that $d_Y(f_N(x),f_N(x_0))<epsilon/3$ if $d_X(x,x_0)<delta$.
This gives us:
$$beginalignd_Y(f(x),f(x_0)&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\
&leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\
&< epsilon/3+epsilon/3+epsilon/3=epsilonendalign$$
Thanks in advance for your correction.
proof-verification metric-spaces continuity
asked Jul 17 at 18:38
Cornman
2,51921128
edited Jul 17 at 18:52
asked Jul 17 at 18:38
Cornman
2,51921128
asked Jul 17 at 18:38
Cornman
2,51921128
asked Jul 17 at 18:38
Cornman
2,51921128
2,51921128
1
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
1
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42
add a comment |Â
1
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
1
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42
1
1
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
1
1
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The idea of this proof is entirely correct, and it's very well executed.
There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<epsilon/3$.
So I would change it to something like this:
Let $xin X$ and $epsilon >0$ be arbitrary. Since $f_nto f$ uniformly, it exists $Nin Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<epsilon/3$ for all $x_1in X$.
As $f_N$ is continuous, there is a $delta>0$ such that for any $x_0$ with $d_X(x, x_0)<delta$ we have $d_Y(f_N(x), f(x_0))<epsilon/3$.
Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $leq$.
answered Jul 17 at 18:49
Arthur
98.8k793175
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
add a comment |Â
up vote
1
down vote
You are close, but in my opinion in one detail you are not sufficiently precise enough.
The problem lies in the inequality
$$
d_Y(f_N(x_0),f(x_0)) leq varepsilon/3,
$$
which you only showed for $x$, not for $x_0$.
In my opinion you should replace the first line of the proof with something like
Let $varepsilon>0$ be arbitrary. Since $f_nto f$ uniformly there exists $Ninmathbb N$ such that $d_Y(f_N(z),f(z))<varepsilon/3$ for all $zin X$.
If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.
answered Jul 17 at 18:49
supinf
5,063926
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The idea of this proof is entirely correct, and it's very well executed.
There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<epsilon/3$.
So I would change it to something like this:
Let $xin X$ and $epsilon >0$ be arbitrary. Since $f_nto f$ uniformly, it exists $Nin Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<epsilon/3$ for all $x_1in X$.
As $f_N$ is continuous, there is a $delta>0$ such that for any $x_0$ with $d_X(x, x_0)<delta$ we have $d_Y(f_N(x), f(x_0))<epsilon/3$.
Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $leq$.
answered Jul 17 at 18:49
Arthur
98.8k793175
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
add a comment |Â
up vote
3
down vote
accepted
The idea of this proof is entirely correct, and it's very well executed.
There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<epsilon/3$.
So I would change it to something like this:
Let $xin X$ and $epsilon >0$ be arbitrary. Since $f_nto f$ uniformly, it exists $Nin Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<epsilon/3$ for all $x_1in X$.
As $f_N$ is continuous, there is a $delta>0$ such that for any $x_0$ with $d_X(x, x_0)<delta$ we have $d_Y(f_N(x), f(x_0))<epsilon/3$.
Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $leq$.
answered Jul 17 at 18:49
Arthur
98.8k793175
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The idea of this proof is entirely correct, and it's very well executed.
There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<epsilon/3$.
So I would change it to something like this:
Let $xin X$ and $epsilon >0$ be arbitrary. Since $f_nto f$ uniformly, it exists $Nin Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<epsilon/3$ for all $x_1in X$.
As $f_N$ is continuous, there is a $delta>0$ such that for any $x_0$ with $d_X(x, x_0)<delta$ we have $d_Y(f_N(x), f(x_0))<epsilon/3$.
Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $leq$.
answered Jul 17 at 18:49
Arthur
98.8k793175
The idea of this proof is entirely correct, and it's very well executed.
There is some minor and entirely inconsequential stuff, like $d_Y(f(x), f(x_0)$ missing a bracket. The biggest thing for me, however, when I read this is that it is not really clear to me where exactly $delta$ comes from. Furthermore, you haven't really made sure that $d_Y(f_N(x_0), f(x_0))<epsilon/3$.
So I would change it to something like this:
Let $xin X$ and $epsilon >0$ be arbitrary. Since $f_nto f$ uniformly, it exists $Nin Bbb N$ such that $d_Y(f_N(x_1),f(x_1))<epsilon/3$ for all $x_1in X$.
As $f_N$ is continuous, there is a $delta>0$ such that for any $x_0$ with $d_X(x, x_0)<delta$ we have $d_Y(f_N(x), f(x_0))<epsilon/3$.
Then from here do as you have already done, except you have to put in at least one $<$ somewhere instead of $leq$.
answered Jul 17 at 18:49
Arthur
98.8k793175
edited Jul 17 at 18:55
answered Jul 17 at 18:49
Arthur
98.8k793175
answered Jul 17 at 18:49
Arthur
98.8k793175
answered Jul 17 at 18:49
Arthur
98.8k793175
98.8k793175
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
add a comment |Â
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
Thank you. I edited that the estimation $d_Y(f_N(x), f(x))<epsilon/3$ holds for every $xin X$. Espacially for $x_0$. This is given by the uniform convergenc.
– Cornman
Jul 17 at 18:55
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
@Cornman That is good. However, I would use a different letter, since you have already fixed an (arbitrary) $x$ and using the same name for two different things is not considered good practice.
– Arthur
Jul 17 at 18:57
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
Ok, I will keep that in mind. :)
– Cornman
Jul 17 at 18:58
add a comment |Â
up vote
1
down vote
You are close, but in my opinion in one detail you are not sufficiently precise enough.
The problem lies in the inequality
$$
d_Y(f_N(x_0),f(x_0)) leq varepsilon/3,
$$
which you only showed for $x$, not for $x_0$.
In my opinion you should replace the first line of the proof with something like
Let $varepsilon>0$ be arbitrary. Since $f_nto f$ uniformly there exists $Ninmathbb N$ such that $d_Y(f_N(z),f(z))<varepsilon/3$ for all $zin X$.
If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.
answered Jul 17 at 18:49
supinf
5,063926
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
add a comment |Â
up vote
1
down vote
You are close, but in my opinion in one detail you are not sufficiently precise enough.
The problem lies in the inequality
$$
d_Y(f_N(x_0),f(x_0)) leq varepsilon/3,
$$
which you only showed for $x$, not for $x_0$.
In my opinion you should replace the first line of the proof with something like
Let $varepsilon>0$ be arbitrary. Since $f_nto f$ uniformly there exists $Ninmathbb N$ such that $d_Y(f_N(z),f(z))<varepsilon/3$ for all $zin X$.
If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.
answered Jul 17 at 18:49
supinf
5,063926
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are close, but in my opinion in one detail you are not sufficiently precise enough.
The problem lies in the inequality
$$
d_Y(f_N(x_0),f(x_0)) leq varepsilon/3,
$$
which you only showed for $x$, not for $x_0$.
In my opinion you should replace the first line of the proof with something like
Let $varepsilon>0$ be arbitrary. Since $f_nto f$ uniformly there exists $Ninmathbb N$ such that $d_Y(f_N(z),f(z))<varepsilon/3$ for all $zin X$.
If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.
answered Jul 17 at 18:49
supinf
5,063926
You are close, but in my opinion in one detail you are not sufficiently precise enough.
The problem lies in the inequality
$$
d_Y(f_N(x_0),f(x_0)) leq varepsilon/3,
$$
which you only showed for $x$, not for $x_0$.
In my opinion you should replace the first line of the proof with something like
Let $varepsilon>0$ be arbitrary. Since $f_nto f$ uniformly there exists $Ninmathbb N$ such that $d_Y(f_N(z),f(z))<varepsilon/3$ for all $zin X$.
If I would read your first line, it sounds like it would also be true if $f_n$ converges pointwise and not uniformly. However, this fact is very important for the proof.
answered Jul 17 at 18:49
supinf
5,063926
answered Jul 17 at 18:49
supinf
5,063926
answered Jul 17 at 18:49
supinf
5,063926
answered Jul 17 at 18:49
supinf
5,063926
5,063926
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
add a comment |Â
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
Thanks for this answer. I forgot to mention, when recapping the definition of uniform convergenc, that this holds for every $xin X$. Espacially for $x_0$. This is were the estimation of $d_Y(f_N(x_0), f(x_0))$ came from.
– Cornman
Jul 17 at 18:54
add a comment |Â
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1
I see no need to correct whatever (other than a typo).
– José Carlos Santos
Jul 17 at 18:41
1
Thanks for the confirmation. Better safe then sorry.
– Cornman
Jul 17 at 18:42