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







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  • Part (1) is only true if the space $X$ is also assumed to complete.
    – s.harp
    20 hours ago















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







share|cite|improve this question







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daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Part (1) is only true if the space $X$ is also assumed to complete.
    – s.harp
    20 hours ago













up vote
0
down vote

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







share|cite|improve this question







New contributor




daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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






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daimengjie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 20 hours ago









daimengjie

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  • Part (1) is only true if the space $X$ is also assumed to complete.
    – s.harp
    20 hours ago


















  • Part (1) is only true if the space $X$ is also assumed to complete.
    – s.harp
    20 hours ago
















Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
20 hours ago




Part (1) is only true if the space $X$ is also assumed to complete.
– s.harp
20 hours ago










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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.






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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.






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      up vote
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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.






      share|cite|improve this answer








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        up vote
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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.






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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.







        share|cite|improve this answer








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        answered 20 hours ago









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