Show that $f$ is a continuous function.
Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$
Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). We want to show that $f(x)$ is continous.
For $x,yin X$ consider
$$|f(x)-f(y)|leq |f(x)-d(a_n,x)|+|d(a_n,x)-d(a_n,y)|+|f(y)-d(a_n,y)|$$
then for large enough $n$ we can make
$$|f(x)-d(a_n,x)|<epsilon/3, |f(y)-d(a_n,y)|<epsilon/3.$$
And since $|d(a_n,x)-d(a_n,y)|<d(x,y)$
we have that
$$|f(x)-f(y)|<2epsilon/3+d(x,y)$$
If we choose $delta = epsilon/3$ then we are done.
Is this proof correct?
real-analysis functional-analysis epsilon-delta
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
add a comment |
Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$
Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). We want to show that $f(x)$ is continous.
For $x,yin X$ consider
$$|f(x)-f(y)|leq |f(x)-d(a_n,x)|+|d(a_n,x)-d(a_n,y)|+|f(y)-d(a_n,y)|$$
then for large enough $n$ we can make
$$|f(x)-d(a_n,x)|<epsilon/3, |f(y)-d(a_n,y)|<epsilon/3.$$
And since $|d(a_n,x)-d(a_n,y)|<d(x,y)$
we have that
$$|f(x)-f(y)|<2epsilon/3+d(x,y)$$
If we choose $delta = epsilon/3$ then we are done.
Is this proof correct?
real-analysis functional-analysis epsilon-delta
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47
add a comment |
Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$
Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). We want to show that $f(x)$ is continous.
For $x,yin X$ consider
$$|f(x)-f(y)|leq |f(x)-d(a_n,x)|+|d(a_n,x)-d(a_n,y)|+|f(y)-d(a_n,y)|$$
then for large enough $n$ we can make
$$|f(x)-d(a_n,x)|<epsilon/3, |f(y)-d(a_n,y)|<epsilon/3.$$
And since $|d(a_n,x)-d(a_n,y)|<d(x,y)$
we have that
$$|f(x)-f(y)|<2epsilon/3+d(x,y)$$
If we choose $delta = epsilon/3$ then we are done.
Is this proof correct?
real-analysis functional-analysis epsilon-delta
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$
Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). We want to show that $f(x)$ is continous.
For $x,yin X$ consider
$$|f(x)-f(y)|leq |f(x)-d(a_n,x)|+|d(a_n,x)-d(a_n,y)|+|f(y)-d(a_n,y)|$$
then for large enough $n$ we can make
$$|f(x)-d(a_n,x)|<epsilon/3, |f(y)-d(a_n,y)|<epsilon/3.$$
And since $|d(a_n,x)-d(a_n,y)|<d(x,y)$
we have that
$$|f(x)-f(y)|<2epsilon/3+d(x,y)$$
If we choose $delta = epsilon/3$ then we are done.
Is this proof correct?
real-analysis functional-analysis epsilon-delta
real-analysis functional-analysis epsilon-delta
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
asked Nov 25 '18 at 17:37
Hello_World
3,87621630
3,87621630
Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47
add a comment |
Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47
Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47
add a comment |
1 Answer
1
active
oldest
votes
There is a faster way:
$$
|,f(x)-f(y)|=lim_{ntoinfty}|d(x,a_n)-d(y,a_n)|le d(x,y).
$$
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
There is a faster way:
$$
|,f(x)-f(y)|=lim_{ntoinfty}|d(x,a_n)-d(y,a_n)|le d(x,y).
$$
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
add a comment |
There is a faster way:
$$
|,f(x)-f(y)|=lim_{ntoinfty}|d(x,a_n)-d(y,a_n)|le d(x,y).
$$
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
add a comment |
There is a faster way:
$$
|,f(x)-f(y)|=lim_{ntoinfty}|d(x,a_n)-d(y,a_n)|le d(x,y).
$$
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
There is a faster way:
$$
|,f(x)-f(y)|=lim_{ntoinfty}|d(x,a_n)-d(y,a_n)|le d(x,y).
$$
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
answered Nov 25 '18 at 17:40
Yiorgos S. Smyrlis
62.6k1383163
62.6k1383163
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
add a comment |
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
Do you think $inf_{xin X} f(x)=0?$
– Hello_World
Nov 25 '18 at 17:54
add a comment |
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Are you assuming the metric space is complete?
– Nicholas Roberts
Nov 25 '18 at 17:41
No information is given about this.
– Hello_World
Nov 25 '18 at 17:43
Nevermind, thats not needed. I notice now that $d(a_n, x)$ will simply be a Cauchy sequence in $mathbb{R}$ and hence convergent.
– Nicholas Roberts
Nov 25 '18 at 17:47