Please check the proof of the following theorem












0















Let $X$ be a metrizable space.If $mathcal A$ is an open covering of
X,then there is an open covering $varepsilon $ of $X$ refining
$mathcal A$ that is countably locally finite.




It is here we will use well-Ordering theorem.Pick a well ordering $<$ of the elements in $mathcal A$.



Choose a metric $d$ for $X$.



Let $n$ be a positive integer,fixed for the moment.



Given an element $U$ of $mathcal A$, let us define $S_n(U)$ as



$$S_n(U) = {x:B(x,frac{1}{n})subset U}$$



For each $Uin mathcal A$, define $$T_n(U) = S_n(U)-bigcup_{V<U} V$$




Now we will show that if $W$ and $V$ are distinct elements of $mathcal{A}$, then $d(T_n(V),T_n(W)) ge frac{1}{n}$.




Assume WLOG that $V < W$, in the linear well-order we chose.



Proof: Let $xin T_n(V)=S_n(V)-cup_{{V'}<V} V'$ which in particular implies that $x in S_n(V)$ and this means that $B(x,frac{1}{n}) subseteq V$.



Let $yin T_n(W)= S_n(W)-cup_{{V'}<W} V'$ which implies that $y notin V$, as it is one of the sets we subtract to form $T_n(W)$ (and we know $y in T_n(W)$ so that $y$ is not in $frac{1}{n}$-neighbourhood of $x$, or it would have been in $V$ by the previous paragraph.



Hence, whenever $xin T_n(V)$ and $yin T_n(W)$, we have $d(x,y)ge frac{1}{n}$.



The sets $T_n(U)$ are not yet the ones we want, for we do not know that they are open sets. So, let us expand each of them slightly to obtain an open set $E_n(U)$. Specifically,let $E_n(U)$ be the $frac{1}{3n}$-neighbourhood of $T_n(U)$. i.e. let $E_n(U)$ be the union of the open balls $B(x,frac{1}{3n})$, for $x in T_n(U)$. Mathematically, $$E_n(U) =bigcup{ B(x,frac{1}{3n}): xin T_n(U) } $$.



($implies T_n(U)subset E_n(U)$)



Let $xin E_n(V),yin E_n(W),vin T_n(V),win T_n(W)$,then $d(v,w)ge frac{1}{n}.$So,$frac{1}{n}le d(v,w)le d(v,x)+d(x,y)+d(y,w)implies d(x,y)ge frac{1}{n}-d(v,x)-d(y,w)ge frac{1}{n}-frac{1}{3n}-frac{1}{3n}=frac{1}{3n}$.



$(d(v,x) le frac{1}{3n})$



So, $ d(E_n(V),E_n(W)) ge frac{1}{3n}$.



let us define: $$varepsilon_n={ E_n(U): U in mathcal A }$$




Claim:$varepsilon_n$ is locally finite collection of open sets that refines $mathcal A$.






  • $varepsilon_n$ refines $mathcal{A}$ comes from the fact that $E_n(V)subset V$ for each $Vin mathcal{A}$.


  • $varepsilon_n$ is locally finite comes from the fact that for any $x in X$, the $frac{1}{6n}$-neighbourhood of $x$ can intersect at most one element of $varepsilon_n$.(1.How to prove this fact?)


  • $varepsilon_n$ will not cover $X$(2.How to prove this fact?).



Claim:$varepsilon=cup_{nin mathbb Z_{+}} varepsilon_n$ covers $X$.




Let $x in X$.
The collection $mathcal A$ with which we began covers $X$.Let us choose $U$ to be the first element of $mathcal{A}$( in the well ordering,$<$)that contains $x$. Since $U$ is open ,we can choose $n$ so that $B(x,frac{1}{n})subset U$. So, $xin S_n(U)$. Now,because $U$ is the first element of $mathcal{A}$ that contains $x$, the point $x in T_n(U)$ implies that $x$ belongs to the element of $E_n(U)$ of $varepsilon_n subseteq varepsilon$. Hence,$varepsilon$ covers $X$ .$blacksquare$










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  • 1




    You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
    – Henno Brandsma
    Nov 24 at 22:25






  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Nov 24 at 22:26










  • @HennoBrandsma:You're right,I forgot to mention that
    – P.Styles
    Nov 24 at 22:26










  • The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
    – Henno Brandsma
    Nov 24 at 22:33












  • What does "countably locally finite" means?
    – Will M.
    Nov 24 at 22:40
















0















Let $X$ be a metrizable space.If $mathcal A$ is an open covering of
X,then there is an open covering $varepsilon $ of $X$ refining
$mathcal A$ that is countably locally finite.




It is here we will use well-Ordering theorem.Pick a well ordering $<$ of the elements in $mathcal A$.



Choose a metric $d$ for $X$.



Let $n$ be a positive integer,fixed for the moment.



Given an element $U$ of $mathcal A$, let us define $S_n(U)$ as



$$S_n(U) = {x:B(x,frac{1}{n})subset U}$$



For each $Uin mathcal A$, define $$T_n(U) = S_n(U)-bigcup_{V<U} V$$




Now we will show that if $W$ and $V$ are distinct elements of $mathcal{A}$, then $d(T_n(V),T_n(W)) ge frac{1}{n}$.




Assume WLOG that $V < W$, in the linear well-order we chose.



Proof: Let $xin T_n(V)=S_n(V)-cup_{{V'}<V} V'$ which in particular implies that $x in S_n(V)$ and this means that $B(x,frac{1}{n}) subseteq V$.



Let $yin T_n(W)= S_n(W)-cup_{{V'}<W} V'$ which implies that $y notin V$, as it is one of the sets we subtract to form $T_n(W)$ (and we know $y in T_n(W)$ so that $y$ is not in $frac{1}{n}$-neighbourhood of $x$, or it would have been in $V$ by the previous paragraph.



Hence, whenever $xin T_n(V)$ and $yin T_n(W)$, we have $d(x,y)ge frac{1}{n}$.



The sets $T_n(U)$ are not yet the ones we want, for we do not know that they are open sets. So, let us expand each of them slightly to obtain an open set $E_n(U)$. Specifically,let $E_n(U)$ be the $frac{1}{3n}$-neighbourhood of $T_n(U)$. i.e. let $E_n(U)$ be the union of the open balls $B(x,frac{1}{3n})$, for $x in T_n(U)$. Mathematically, $$E_n(U) =bigcup{ B(x,frac{1}{3n}): xin T_n(U) } $$.



($implies T_n(U)subset E_n(U)$)



Let $xin E_n(V),yin E_n(W),vin T_n(V),win T_n(W)$,then $d(v,w)ge frac{1}{n}.$So,$frac{1}{n}le d(v,w)le d(v,x)+d(x,y)+d(y,w)implies d(x,y)ge frac{1}{n}-d(v,x)-d(y,w)ge frac{1}{n}-frac{1}{3n}-frac{1}{3n}=frac{1}{3n}$.



$(d(v,x) le frac{1}{3n})$



So, $ d(E_n(V),E_n(W)) ge frac{1}{3n}$.



let us define: $$varepsilon_n={ E_n(U): U in mathcal A }$$




Claim:$varepsilon_n$ is locally finite collection of open sets that refines $mathcal A$.






  • $varepsilon_n$ refines $mathcal{A}$ comes from the fact that $E_n(V)subset V$ for each $Vin mathcal{A}$.


  • $varepsilon_n$ is locally finite comes from the fact that for any $x in X$, the $frac{1}{6n}$-neighbourhood of $x$ can intersect at most one element of $varepsilon_n$.(1.How to prove this fact?)


  • $varepsilon_n$ will not cover $X$(2.How to prove this fact?).



Claim:$varepsilon=cup_{nin mathbb Z_{+}} varepsilon_n$ covers $X$.




Let $x in X$.
The collection $mathcal A$ with which we began covers $X$.Let us choose $U$ to be the first element of $mathcal{A}$( in the well ordering,$<$)that contains $x$. Since $U$ is open ,we can choose $n$ so that $B(x,frac{1}{n})subset U$. So, $xin S_n(U)$. Now,because $U$ is the first element of $mathcal{A}$ that contains $x$, the point $x in T_n(U)$ implies that $x$ belongs to the element of $E_n(U)$ of $varepsilon_n subseteq varepsilon$. Hence,$varepsilon$ covers $X$ .$blacksquare$










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  • 1




    You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
    – Henno Brandsma
    Nov 24 at 22:25






  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Nov 24 at 22:26










  • @HennoBrandsma:You're right,I forgot to mention that
    – P.Styles
    Nov 24 at 22:26










  • The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
    – Henno Brandsma
    Nov 24 at 22:33












  • What does "countably locally finite" means?
    – Will M.
    Nov 24 at 22:40














0












0








0








Let $X$ be a metrizable space.If $mathcal A$ is an open covering of
X,then there is an open covering $varepsilon $ of $X$ refining
$mathcal A$ that is countably locally finite.




It is here we will use well-Ordering theorem.Pick a well ordering $<$ of the elements in $mathcal A$.



Choose a metric $d$ for $X$.



Let $n$ be a positive integer,fixed for the moment.



Given an element $U$ of $mathcal A$, let us define $S_n(U)$ as



$$S_n(U) = {x:B(x,frac{1}{n})subset U}$$



For each $Uin mathcal A$, define $$T_n(U) = S_n(U)-bigcup_{V<U} V$$




Now we will show that if $W$ and $V$ are distinct elements of $mathcal{A}$, then $d(T_n(V),T_n(W)) ge frac{1}{n}$.




Assume WLOG that $V < W$, in the linear well-order we chose.



Proof: Let $xin T_n(V)=S_n(V)-cup_{{V'}<V} V'$ which in particular implies that $x in S_n(V)$ and this means that $B(x,frac{1}{n}) subseteq V$.



Let $yin T_n(W)= S_n(W)-cup_{{V'}<W} V'$ which implies that $y notin V$, as it is one of the sets we subtract to form $T_n(W)$ (and we know $y in T_n(W)$ so that $y$ is not in $frac{1}{n}$-neighbourhood of $x$, or it would have been in $V$ by the previous paragraph.



Hence, whenever $xin T_n(V)$ and $yin T_n(W)$, we have $d(x,y)ge frac{1}{n}$.



The sets $T_n(U)$ are not yet the ones we want, for we do not know that they are open sets. So, let us expand each of them slightly to obtain an open set $E_n(U)$. Specifically,let $E_n(U)$ be the $frac{1}{3n}$-neighbourhood of $T_n(U)$. i.e. let $E_n(U)$ be the union of the open balls $B(x,frac{1}{3n})$, for $x in T_n(U)$. Mathematically, $$E_n(U) =bigcup{ B(x,frac{1}{3n}): xin T_n(U) } $$.



($implies T_n(U)subset E_n(U)$)



Let $xin E_n(V),yin E_n(W),vin T_n(V),win T_n(W)$,then $d(v,w)ge frac{1}{n}.$So,$frac{1}{n}le d(v,w)le d(v,x)+d(x,y)+d(y,w)implies d(x,y)ge frac{1}{n}-d(v,x)-d(y,w)ge frac{1}{n}-frac{1}{3n}-frac{1}{3n}=frac{1}{3n}$.



$(d(v,x) le frac{1}{3n})$



So, $ d(E_n(V),E_n(W)) ge frac{1}{3n}$.



let us define: $$varepsilon_n={ E_n(U): U in mathcal A }$$




Claim:$varepsilon_n$ is locally finite collection of open sets that refines $mathcal A$.






  • $varepsilon_n$ refines $mathcal{A}$ comes from the fact that $E_n(V)subset V$ for each $Vin mathcal{A}$.


  • $varepsilon_n$ is locally finite comes from the fact that for any $x in X$, the $frac{1}{6n}$-neighbourhood of $x$ can intersect at most one element of $varepsilon_n$.(1.How to prove this fact?)


  • $varepsilon_n$ will not cover $X$(2.How to prove this fact?).



Claim:$varepsilon=cup_{nin mathbb Z_{+}} varepsilon_n$ covers $X$.




Let $x in X$.
The collection $mathcal A$ with which we began covers $X$.Let us choose $U$ to be the first element of $mathcal{A}$( in the well ordering,$<$)that contains $x$. Since $U$ is open ,we can choose $n$ so that $B(x,frac{1}{n})subset U$. So, $xin S_n(U)$. Now,because $U$ is the first element of $mathcal{A}$ that contains $x$, the point $x in T_n(U)$ implies that $x$ belongs to the element of $E_n(U)$ of $varepsilon_n subseteq varepsilon$. Hence,$varepsilon$ covers $X$ .$blacksquare$










share|cite|improve this question
















Let $X$ be a metrizable space.If $mathcal A$ is an open covering of
X,then there is an open covering $varepsilon $ of $X$ refining
$mathcal A$ that is countably locally finite.




It is here we will use well-Ordering theorem.Pick a well ordering $<$ of the elements in $mathcal A$.



Choose a metric $d$ for $X$.



Let $n$ be a positive integer,fixed for the moment.



Given an element $U$ of $mathcal A$, let us define $S_n(U)$ as



$$S_n(U) = {x:B(x,frac{1}{n})subset U}$$



For each $Uin mathcal A$, define $$T_n(U) = S_n(U)-bigcup_{V<U} V$$




Now we will show that if $W$ and $V$ are distinct elements of $mathcal{A}$, then $d(T_n(V),T_n(W)) ge frac{1}{n}$.




Assume WLOG that $V < W$, in the linear well-order we chose.



Proof: Let $xin T_n(V)=S_n(V)-cup_{{V'}<V} V'$ which in particular implies that $x in S_n(V)$ and this means that $B(x,frac{1}{n}) subseteq V$.



Let $yin T_n(W)= S_n(W)-cup_{{V'}<W} V'$ which implies that $y notin V$, as it is one of the sets we subtract to form $T_n(W)$ (and we know $y in T_n(W)$ so that $y$ is not in $frac{1}{n}$-neighbourhood of $x$, or it would have been in $V$ by the previous paragraph.



Hence, whenever $xin T_n(V)$ and $yin T_n(W)$, we have $d(x,y)ge frac{1}{n}$.



The sets $T_n(U)$ are not yet the ones we want, for we do not know that they are open sets. So, let us expand each of them slightly to obtain an open set $E_n(U)$. Specifically,let $E_n(U)$ be the $frac{1}{3n}$-neighbourhood of $T_n(U)$. i.e. let $E_n(U)$ be the union of the open balls $B(x,frac{1}{3n})$, for $x in T_n(U)$. Mathematically, $$E_n(U) =bigcup{ B(x,frac{1}{3n}): xin T_n(U) } $$.



($implies T_n(U)subset E_n(U)$)



Let $xin E_n(V),yin E_n(W),vin T_n(V),win T_n(W)$,then $d(v,w)ge frac{1}{n}.$So,$frac{1}{n}le d(v,w)le d(v,x)+d(x,y)+d(y,w)implies d(x,y)ge frac{1}{n}-d(v,x)-d(y,w)ge frac{1}{n}-frac{1}{3n}-frac{1}{3n}=frac{1}{3n}$.



$(d(v,x) le frac{1}{3n})$



So, $ d(E_n(V),E_n(W)) ge frac{1}{3n}$.



let us define: $$varepsilon_n={ E_n(U): U in mathcal A }$$




Claim:$varepsilon_n$ is locally finite collection of open sets that refines $mathcal A$.






  • $varepsilon_n$ refines $mathcal{A}$ comes from the fact that $E_n(V)subset V$ for each $Vin mathcal{A}$.


  • $varepsilon_n$ is locally finite comes from the fact that for any $x in X$, the $frac{1}{6n}$-neighbourhood of $x$ can intersect at most one element of $varepsilon_n$.(1.How to prove this fact?)


  • $varepsilon_n$ will not cover $X$(2.How to prove this fact?).



Claim:$varepsilon=cup_{nin mathbb Z_{+}} varepsilon_n$ covers $X$.




Let $x in X$.
The collection $mathcal A$ with which we began covers $X$.Let us choose $U$ to be the first element of $mathcal{A}$( in the well ordering,$<$)that contains $x$. Since $U$ is open ,we can choose $n$ so that $B(x,frac{1}{n})subset U$. So, $xin S_n(U)$. Now,because $U$ is the first element of $mathcal{A}$ that contains $x$, the point $x in T_n(U)$ implies that $x$ belongs to the element of $E_n(U)$ of $varepsilon_n subseteq varepsilon$. Hence,$varepsilon$ covers $X$ .$blacksquare$







real-analysis general-topology proof-verification proof-explanation






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edited yesterday

























asked Nov 24 at 22:21









P.Styles

1,420726




1,420726








  • 1




    You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
    – Henno Brandsma
    Nov 24 at 22:25






  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Nov 24 at 22:26










  • @HennoBrandsma:You're right,I forgot to mention that
    – P.Styles
    Nov 24 at 22:26










  • The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
    – Henno Brandsma
    Nov 24 at 22:33












  • What does "countably locally finite" means?
    – Will M.
    Nov 24 at 22:40














  • 1




    You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
    – Henno Brandsma
    Nov 24 at 22:25






  • 1




    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    – Shaun
    Nov 24 at 22:26










  • @HennoBrandsma:You're right,I forgot to mention that
    – P.Styles
    Nov 24 at 22:26










  • The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
    – Henno Brandsma
    Nov 24 at 22:33












  • What does "countably locally finite" means?
    – Will M.
    Nov 24 at 22:40








1




1




You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
– Henno Brandsma
Nov 24 at 22:25




You end a bit abruptly.... Also, you have chosen an order on $mathcal{A}$ that you use in the definition of the $T_n(U)$? a well-order perhaps?
– Henno Brandsma
Nov 24 at 22:25




1




1




Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Nov 24 at 22:26




Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Nov 24 at 22:26












@HennoBrandsma:You're right,I forgot to mention that
– P.Styles
Nov 24 at 22:26




@HennoBrandsma:You're right,I forgot to mention that
– P.Styles
Nov 24 at 22:26












The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
– Henno Brandsma
Nov 24 at 22:33






The fact that $epsilon_n$ does not (need to) cover $X$ need not be shown at all. The $K_1$ notation is "random", don't you mean $V$ as in the definition??
– Henno Brandsma
Nov 24 at 22:33














What does "countably locally finite" means?
– Will M.
Nov 24 at 22:40




What does "countably locally finite" means?
– Will M.
Nov 24 at 22:40










1 Answer
1






active

oldest

votes


















2















  • You should start out by explicitly chosing a well-order $<$ on the open cover $mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).


  • The remark that the $varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.


  • I find the proof that $d(E_n(V),E_n(W)) ge frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)


  • To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) ge frac{1}{3n}$ whenever $U neq V$. If a ball $B(x, frac{1}{6n})$ would intersect two members of $varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that
    $$d(p,q) le d(p,x) + d(q,x) < frac{1}{6n} + frac{1}{6n} = frac{1}{3n}$$ and
    thus $d(E_n(V), E_n(V)) < frac{1}{3n}$ contradicting what we already knew. So every $x in X$ has a neighbourhood $B(x, frac{1}{6n})$ that intersects at most one member of $varepsilon_n$ (so $varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $varepsilon$).


  • You could explain $E_n(V) subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.


  • You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.



This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.






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  • :Please see the edit in response to your third suggestion.
    – P.Styles
    yesterday













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  • You should start out by explicitly chosing a well-order $<$ on the open cover $mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).


  • The remark that the $varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.


  • I find the proof that $d(E_n(V),E_n(W)) ge frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)


  • To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) ge frac{1}{3n}$ whenever $U neq V$. If a ball $B(x, frac{1}{6n})$ would intersect two members of $varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that
    $$d(p,q) le d(p,x) + d(q,x) < frac{1}{6n} + frac{1}{6n} = frac{1}{3n}$$ and
    thus $d(E_n(V), E_n(V)) < frac{1}{3n}$ contradicting what we already knew. So every $x in X$ has a neighbourhood $B(x, frac{1}{6n})$ that intersects at most one member of $varepsilon_n$ (so $varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $varepsilon$).


  • You could explain $E_n(V) subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.


  • You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.



This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.






share|cite|improve this answer























  • :Please see the edit in response to your third suggestion.
    – P.Styles
    yesterday


















2















  • You should start out by explicitly chosing a well-order $<$ on the open cover $mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).


  • The remark that the $varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.


  • I find the proof that $d(E_n(V),E_n(W)) ge frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)


  • To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) ge frac{1}{3n}$ whenever $U neq V$. If a ball $B(x, frac{1}{6n})$ would intersect two members of $varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that
    $$d(p,q) le d(p,x) + d(q,x) < frac{1}{6n} + frac{1}{6n} = frac{1}{3n}$$ and
    thus $d(E_n(V), E_n(V)) < frac{1}{3n}$ contradicting what we already knew. So every $x in X$ has a neighbourhood $B(x, frac{1}{6n})$ that intersects at most one member of $varepsilon_n$ (so $varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $varepsilon$).


  • You could explain $E_n(V) subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.


  • You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.



This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.






share|cite|improve this answer























  • :Please see the edit in response to your third suggestion.
    – P.Styles
    yesterday
















2












2








2







  • You should start out by explicitly chosing a well-order $<$ on the open cover $mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).


  • The remark that the $varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.


  • I find the proof that $d(E_n(V),E_n(W)) ge frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)


  • To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) ge frac{1}{3n}$ whenever $U neq V$. If a ball $B(x, frac{1}{6n})$ would intersect two members of $varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that
    $$d(p,q) le d(p,x) + d(q,x) < frac{1}{6n} + frac{1}{6n} = frac{1}{3n}$$ and
    thus $d(E_n(V), E_n(V)) < frac{1}{3n}$ contradicting what we already knew. So every $x in X$ has a neighbourhood $B(x, frac{1}{6n})$ that intersects at most one member of $varepsilon_n$ (so $varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $varepsilon$).


  • You could explain $E_n(V) subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.


  • You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.



This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.






share|cite|improve this answer















  • You should start out by explicitly chosing a well-order $<$ on the open cover $mathcal{A}$. This hints (which is true) that the theorem essentially uses the axiom of choice (which I personally have no problem with, but some people might).


  • The remark that the $varepsilon_n$ need not cover $X$ is just pedagogical. What only matters is that the union of all these does cover $X$, which is your final point.


  • I find the proof that $d(E_n(V),E_n(W)) ge frac{1}{3n}$ unclearly written. Try and improve that (how do you introduce your variables e.g.)


  • To see your $1$, use what was shown above it: namely that $d(E_n(V), E_n(V)) ge frac{1}{3n}$ whenever $U neq V$. If a ball $B(x, frac{1}{6n})$ would intersect two members of $varepsilon_n$, it would intersect say $E_n(V)$ (in $p$ say) and $E_n(W)$ (say in $q$) and then computing the distance between these intersection points via $x$, shows that
    $$d(p,q) le d(p,x) + d(q,x) < frac{1}{6n} + frac{1}{6n} = frac{1}{3n}$$ and
    thus $d(E_n(V), E_n(V)) < frac{1}{3n}$ contradicting what we already knew. So every $x in X$ has a neighbourhood $B(x, frac{1}{6n})$ that intersects at most one member of $varepsilon_n$ (so $varepsilon_n$ is even discrete, not just locally finite, and we have a countably discrete cover $varepsilon$).


  • You could explain $E_n(V) subseteq V$ better as well. Why does that hold? It's true but a proof should explain it.


  • You could end with "and this concludes the proof", or quod erat demonstrandum or ὅπερ ἔδει δεῖξαι etc. instead of so abruptly.



This proof is essentially the same one as given in Munkres (2nd edition), p246-247, so no credits for originality.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 23:46

























answered Nov 24 at 22:59









Henno Brandsma

104k346113




104k346113












  • :Please see the edit in response to your third suggestion.
    – P.Styles
    yesterday




















  • :Please see the edit in response to your third suggestion.
    – P.Styles
    yesterday


















:Please see the edit in response to your third suggestion.
– P.Styles
yesterday






:Please see the edit in response to your third suggestion.
– P.Styles
yesterday




















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