Question about proof “ every perfect set is uncountable.”.

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This is my proof refer to baby rudin Thm 2.43.
Let $P$ is perfect set, then $P$ is infinite set.(Because P don't have any isolated point.)
If $P$ is countable $P$ should be represented by
$$P=x_1,x_2,cdots $$
Now labeling following way, Because P is non-empty
,$exists x_1in P$,



Let a arbitrary neighborhood $N_r_1(x_1)$.



Since every point in $P$ is limit point of $P$.
$exists x_2 in P$.



Let $r_2=frac12min(d(x_1,x_2),r_1-d(x_1,x_2))$



Then $(i)barN_r_2(x_2)subset N_r_1(x_1)$, ($barN$ imply closure of N)
,$(ii) x_1notin N_r_2(x_2)$,$(iii)N_r_2(x_2)cap Pneq emptyset$
similar way we can make sequence $N_r_n(x_n)$ satisfying



$(i)barN_r_n+1(x_n+1)subset N_r_n(x_n)$
,$(ii) x_nnotin N_r_n+1(x_n+1)$,$(iii)N_r_n+1(x_n+1)cap Pneq emptyset$



Let $K_n:=barN_r_n(x_n)cap P$, then P is closed set, and $barN_r_n(x_n)$ is closed and bounded, so $K_n$ is closed subset of compact set, therefore, $K_n$ is compact.



$forall nin N$, $K_n$ is non-empty compact set satisfying $K_n+1in K_n$,
so $bigcap_nin NK_nneq emptyset.$
Otherwise, $x_nnotin K_n+1$, so there should be exist $yneq x_n, forall nin N$, $yinbigcap_nin NK_n$. $P$ is not represented by $x_n$, so P is uncountable.



Here is question,
I don't know what is difference between above proof and following,



For $x_1in N$, every sequence,$x_n+1=x_n+2$ is not represented $N$ perfectly, so $N$ is uncountable.



what is the difference of two proofs?, or If my proof for" every perfect set is uncountable." is not exact, where is exact proof?










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  • I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
    – Aniruddha Deshmukh
    Aug 31 at 6:23










  • This is my fault,
    – ë°±ì£¼ìƒ
    Aug 31 at 6:27










  • Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
    – Aniruddha Deshmukh
    Aug 31 at 6:31










  • Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
    – Aniruddha Deshmukh
    Aug 31 at 6:33










  • According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
    – ë°±ì£¼ìƒ
    Aug 31 at 6:35














up vote
2
down vote

favorite
2












This is my proof refer to baby rudin Thm 2.43.
Let $P$ is perfect set, then $P$ is infinite set.(Because P don't have any isolated point.)
If $P$ is countable $P$ should be represented by
$$P=x_1,x_2,cdots $$
Now labeling following way, Because P is non-empty
,$exists x_1in P$,



Let a arbitrary neighborhood $N_r_1(x_1)$.



Since every point in $P$ is limit point of $P$.
$exists x_2 in P$.



Let $r_2=frac12min(d(x_1,x_2),r_1-d(x_1,x_2))$



Then $(i)barN_r_2(x_2)subset N_r_1(x_1)$, ($barN$ imply closure of N)
,$(ii) x_1notin N_r_2(x_2)$,$(iii)N_r_2(x_2)cap Pneq emptyset$
similar way we can make sequence $N_r_n(x_n)$ satisfying



$(i)barN_r_n+1(x_n+1)subset N_r_n(x_n)$
,$(ii) x_nnotin N_r_n+1(x_n+1)$,$(iii)N_r_n+1(x_n+1)cap Pneq emptyset$



Let $K_n:=barN_r_n(x_n)cap P$, then P is closed set, and $barN_r_n(x_n)$ is closed and bounded, so $K_n$ is closed subset of compact set, therefore, $K_n$ is compact.



$forall nin N$, $K_n$ is non-empty compact set satisfying $K_n+1in K_n$,
so $bigcap_nin NK_nneq emptyset.$
Otherwise, $x_nnotin K_n+1$, so there should be exist $yneq x_n, forall nin N$, $yinbigcap_nin NK_n$. $P$ is not represented by $x_n$, so P is uncountable.



Here is question,
I don't know what is difference between above proof and following,



For $x_1in N$, every sequence,$x_n+1=x_n+2$ is not represented $N$ perfectly, so $N$ is uncountable.



what is the difference of two proofs?, or If my proof for" every perfect set is uncountable." is not exact, where is exact proof?










share|cite|improve this question























  • I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
    – Aniruddha Deshmukh
    Aug 31 at 6:23










  • This is my fault,
    – ë°±ì£¼ìƒ
    Aug 31 at 6:27










  • Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
    – Aniruddha Deshmukh
    Aug 31 at 6:31










  • Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
    – Aniruddha Deshmukh
    Aug 31 at 6:33










  • According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
    – ë°±ì£¼ìƒ
    Aug 31 at 6:35












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





This is my proof refer to baby rudin Thm 2.43.
Let $P$ is perfect set, then $P$ is infinite set.(Because P don't have any isolated point.)
If $P$ is countable $P$ should be represented by
$$P=x_1,x_2,cdots $$
Now labeling following way, Because P is non-empty
,$exists x_1in P$,



Let a arbitrary neighborhood $N_r_1(x_1)$.



Since every point in $P$ is limit point of $P$.
$exists x_2 in P$.



Let $r_2=frac12min(d(x_1,x_2),r_1-d(x_1,x_2))$



Then $(i)barN_r_2(x_2)subset N_r_1(x_1)$, ($barN$ imply closure of N)
,$(ii) x_1notin N_r_2(x_2)$,$(iii)N_r_2(x_2)cap Pneq emptyset$
similar way we can make sequence $N_r_n(x_n)$ satisfying



$(i)barN_r_n+1(x_n+1)subset N_r_n(x_n)$
,$(ii) x_nnotin N_r_n+1(x_n+1)$,$(iii)N_r_n+1(x_n+1)cap Pneq emptyset$



Let $K_n:=barN_r_n(x_n)cap P$, then P is closed set, and $barN_r_n(x_n)$ is closed and bounded, so $K_n$ is closed subset of compact set, therefore, $K_n$ is compact.



$forall nin N$, $K_n$ is non-empty compact set satisfying $K_n+1in K_n$,
so $bigcap_nin NK_nneq emptyset.$
Otherwise, $x_nnotin K_n+1$, so there should be exist $yneq x_n, forall nin N$, $yinbigcap_nin NK_n$. $P$ is not represented by $x_n$, so P is uncountable.



Here is question,
I don't know what is difference between above proof and following,



For $x_1in N$, every sequence,$x_n+1=x_n+2$ is not represented $N$ perfectly, so $N$ is uncountable.



what is the difference of two proofs?, or If my proof for" every perfect set is uncountable." is not exact, where is exact proof?










share|cite|improve this question















This is my proof refer to baby rudin Thm 2.43.
Let $P$ is perfect set, then $P$ is infinite set.(Because P don't have any isolated point.)
If $P$ is countable $P$ should be represented by
$$P=x_1,x_2,cdots $$
Now labeling following way, Because P is non-empty
,$exists x_1in P$,



Let a arbitrary neighborhood $N_r_1(x_1)$.



Since every point in $P$ is limit point of $P$.
$exists x_2 in P$.



Let $r_2=frac12min(d(x_1,x_2),r_1-d(x_1,x_2))$



Then $(i)barN_r_2(x_2)subset N_r_1(x_1)$, ($barN$ imply closure of N)
,$(ii) x_1notin N_r_2(x_2)$,$(iii)N_r_2(x_2)cap Pneq emptyset$
similar way we can make sequence $N_r_n(x_n)$ satisfying



$(i)barN_r_n+1(x_n+1)subset N_r_n(x_n)$
,$(ii) x_nnotin N_r_n+1(x_n+1)$,$(iii)N_r_n+1(x_n+1)cap Pneq emptyset$



Let $K_n:=barN_r_n(x_n)cap P$, then P is closed set, and $barN_r_n(x_n)$ is closed and bounded, so $K_n$ is closed subset of compact set, therefore, $K_n$ is compact.



$forall nin N$, $K_n$ is non-empty compact set satisfying $K_n+1in K_n$,
so $bigcap_nin NK_nneq emptyset.$
Otherwise, $x_nnotin K_n+1$, so there should be exist $yneq x_n, forall nin N$, $yinbigcap_nin NK_n$. $P$ is not represented by $x_n$, so P is uncountable.



Here is question,
I don't know what is difference between above proof and following,



For $x_1in N$, every sequence,$x_n+1=x_n+2$ is not represented $N$ perfectly, so $N$ is uncountable.



what is the difference of two proofs?, or If my proof for" every perfect set is uncountable." is not exact, where is exact proof?







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share|cite|improve this question













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edited Aug 31 at 6:27

























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  • I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
    – Aniruddha Deshmukh
    Aug 31 at 6:23










  • This is my fault,
    – ë°±ì£¼ìƒ
    Aug 31 at 6:27










  • Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
    – Aniruddha Deshmukh
    Aug 31 at 6:31










  • Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
    – Aniruddha Deshmukh
    Aug 31 at 6:33










  • According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
    – ë°±ì£¼ìƒ
    Aug 31 at 6:35
















  • I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
    – Aniruddha Deshmukh
    Aug 31 at 6:23










  • This is my fault,
    – ë°±ì£¼ìƒ
    Aug 31 at 6:27










  • Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
    – Aniruddha Deshmukh
    Aug 31 at 6:31










  • Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
    – Aniruddha Deshmukh
    Aug 31 at 6:33










  • According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
    – ë°±ì£¼ìƒ
    Aug 31 at 6:35















I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
– Aniruddha Deshmukh
Aug 31 at 6:23




I think you can conclude that a perfect set is infinite because it has no isolated points. Not because it is closed. Because every finite subset of $mathbbR$ is closed, but it is not perfect. At least, as far as I have read about perfect sets.
– Aniruddha Deshmukh
Aug 31 at 6:23












This is my fault,
– ë°±ì£¼ìƒ
Aug 31 at 6:27




This is my fault,
– ë°±ì£¼ìƒ
Aug 31 at 6:27












Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
– Aniruddha Deshmukh
Aug 31 at 6:31




Also, how can you conclude (i)? Suppose your $P = left[ 0, 1 right]$. Take $x_1 = dfrac14$ with $r_1 = dfrac11000$ and then take $x_2 = dfrac34$, then you will not have the closure of neighbourhood of $x_2$ as a subset of neighbourhood of $x_1$.
– Aniruddha Deshmukh
Aug 31 at 6:31












Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
– Aniruddha Deshmukh
Aug 31 at 6:33




Also, I am thinking of using the Hausdroff property of metric spaces to get a contradiction. However, the infimum would create a problem when constructing a neighburhood to prove a point to be isolated.
– Aniruddha Deshmukh
Aug 31 at 6:33












According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
– ë°±ì£¼ìƒ
Aug 31 at 6:35




According to my proof, we should select $x_2$ in $[frac14-frac12000,frac14+frac12000]$.
– ë°±ì£¼ìƒ
Aug 31 at 6:35















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