A functional analysis exam question
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Let $X$ be the metric space and it is not a compact set.Show that
$(1)$There is $varepsilon>0$ and the sequence $left{ x_n right}subset X$ ,when $mne n$,there is$$Bleft( x_n,varepsilon right) cap Bleft( x_m,varepsilon right)=oslash.$$
$(2)$There is a continuous function $f_n(x):Xlongrightarrow left[ text{0,}1 right]$ for any $n$,such that
$$f_n(x_{n})=1$$if and only if $xnotin Bleft( x,frac{varepsilon}{2} right)$,there is$f_n(x)=0.$
I worked hard but didn't solve it.I started from a definition that is not compact set, but I don't know how to find the sequence $left{ x_n right}$.So I hope you can give me some ideas.
functional-analysis
asked Nov 19 at 10:58
daimengjie
6
add a comment |
up vote
0
down vote
favorite
Let $X$ be the metric space and it is not a compact set.Show that
$(1)$There is $varepsilon>0$ and the sequence $left{ x_n right}subset X$ ,when $mne n$,there is$$Bleft( x_n,varepsilon right) cap Bleft( x_m,varepsilon right)=oslash.$$
$(2)$There is a continuous function $f_n(x):Xlongrightarrow left[ text{0,}1 right]$ for any $n$,such that
$$f_n(x_{n})=1$$if and only if $xnotin Bleft( x,frac{varepsilon}{2} right)$,there is$f_n(x)=0.$
I worked hard but didn't solve it.I started from a definition that is not compact set, but I don't know how to find the sequence $left{ x_n right}$.So I hope you can give me some ideas.
functional-analysis
asked Nov 19 at 10:58
daimengjie
6
Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09
add a comment |
up vote
0
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favorite
up vote
0
down vote
favorite
Let $X$ be the metric space and it is not a compact set.Show that
$(1)$There is $varepsilon>0$ and the sequence $left{ x_n right}subset X$ ,when $mne n$,there is$$Bleft( x_n,varepsilon right) cap Bleft( x_m,varepsilon right)=oslash.$$
$(2)$There is a continuous function $f_n(x):Xlongrightarrow left[ text{0,}1 right]$ for any $n$,such that
$$f_n(x_{n})=1$$if and only if $xnotin Bleft( x,frac{varepsilon}{2} right)$,there is$f_n(x)=0.$
I worked hard but didn't solve it.I started from a definition that is not compact set, but I don't know how to find the sequence $left{ x_n right}$.So I hope you can give me some ideas.
functional-analysis
asked Nov 19 at 10:58
daimengjie
6
Let $X$ be the metric space and it is not a compact set.Show that
$(1)$There is $varepsilon>0$ and the sequence $left{ x_n right}subset X$ ,when $mne n$,there is$$Bleft( x_n,varepsilon right) cap Bleft( x_m,varepsilon right)=oslash.$$
$(2)$There is a continuous function $f_n(x):Xlongrightarrow left[ text{0,}1 right]$ for any $n$,such that
$$f_n(x_{n})=1$$if and only if $xnotin Bleft( x,frac{varepsilon}{2} right)$,there is$f_n(x)=0.$
I worked hard but didn't solve it.I started from a definition that is not compact set, but I don't know how to find the sequence $left{ x_n right}$.So I hope you can give me some ideas.
functional-analysis
functional-analysis
asked Nov 19 at 10:58
daimengjie
6
asked Nov 19 at 10:58
daimengjie
6
asked Nov 19 at 10:58
daimengjie
6
asked Nov 19 at 10:58
daimengjie
6
asked Nov 19 at 10:58
daimengjie
6
6
Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09
add a comment |
Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09
Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09
Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09
add a comment |
1 Answer
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You need extra hypothesis for $(1)$. For example consider the open interval $(0,1)$. Because the mention to 'functional analysis' in the title I suppose that $X$ is an infinite dimensional normed space or similar.
answered Nov 19 at 11:08
Dante Grevino
7817
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
up vote
0
down vote
You need extra hypothesis for $(1)$. For example consider the open interval $(0,1)$. Because the mention to 'functional analysis' in the title I suppose that $X$ is an infinite dimensional normed space or similar.
answered Nov 19 at 11:08
Dante Grevino
7817
add a comment |
up vote
0
down vote
You need extra hypothesis for $(1)$. For example consider the open interval $(0,1)$. Because the mention to 'functional analysis' in the title I suppose that $X$ is an infinite dimensional normed space or similar.
answered Nov 19 at 11:08
Dante Grevino
7817
add a comment |
up vote
0
down vote
up vote
0
down vote
You need extra hypothesis for $(1)$. For example consider the open interval $(0,1)$. Because the mention to 'functional analysis' in the title I suppose that $X$ is an infinite dimensional normed space or similar.
answered Nov 19 at 11:08
Dante Grevino
7817
You need extra hypothesis for $(1)$. For example consider the open interval $(0,1)$. Because the mention to 'functional analysis' in the title I suppose that $X$ is an infinite dimensional normed space or similar.
answered Nov 19 at 11:08
Dante Grevino
7817
answered Nov 19 at 11:08
Dante Grevino
7817
answered Nov 19 at 11:08
Dante Grevino
7817
answered Nov 19 at 11:08
Dante Grevino
7817
7817
add a comment |
add a comment |
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Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
Nov 19 at 11:09