Intersection of finite collection of open sets is open











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Can someone help me through this theorem in 'Baby Rudin.'



For any collection $G_{1},...,G_{n}$ of open sets, $bigcaplimits_{i = 1}^{n} G_{i}$ is open.



Pf
Put $H = bigcaplimits_{i=1}^{n} G_{i}$. For any $x in H$, there exist neighborhoods $N_{i}$ of $x$, with radii $r_{i}$, such that $N_{i} subset G_{i} (i = 1,...,n)$. Put $$r = min(r_{1},...,r_{n})$$ and let $N$ be the neighborhood of $x$ of radius $r$. Then $N subset G_{i}$ for $i = 1,...,n$, so that $N subset H$, and $H$ is open.



$Box$



So I am having difficulty in seeing why we are able to say the neighborhood $N_{r}(x)$ is a subset of $G_{i}$ for every $i$. I am not following here. Any ideas?



Thanks.






analysis







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  • The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
    – user122283
    May 28 '14 at 14:07










  • Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
    – user122283
    May 28 '14 at 14:10










  • Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
    – zzz2991
    May 28 '14 at 15:05















up vote
0
down vote

favorite












Can someone help me through this theorem in 'Baby Rudin.'



For any collection $G_{1},...,G_{n}$ of open sets, $bigcaplimits_{i = 1}^{n} G_{i}$ is open.



Pf
Put $H = bigcaplimits_{i=1}^{n} G_{i}$. For any $x in H$, there exist neighborhoods $N_{i}$ of $x$, with radii $r_{i}$, such that $N_{i} subset G_{i} (i = 1,...,n)$. Put $$r = min(r_{1},...,r_{n})$$ and let $N$ be the neighborhood of $x$ of radius $r$. Then $N subset G_{i}$ for $i = 1,...,n$, so that $N subset H$, and $H$ is open.



$Box$



So I am having difficulty in seeing why we are able to say the neighborhood $N_{r}(x)$ is a subset of $G_{i}$ for every $i$. I am not following here. Any ideas?



Thanks.






analysis







share|cite|improve this question
























  • The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
    – user122283
    May 28 '14 at 14:07










  • Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
    – user122283
    May 28 '14 at 14:10










  • Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
    – zzz2991
    May 28 '14 at 15:05













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Can someone help me through this theorem in 'Baby Rudin.'



For any collection $G_{1},...,G_{n}$ of open sets, $bigcaplimits_{i = 1}^{n} G_{i}$ is open.



Pf
Put $H = bigcaplimits_{i=1}^{n} G_{i}$. For any $x in H$, there exist neighborhoods $N_{i}$ of $x$, with radii $r_{i}$, such that $N_{i} subset G_{i} (i = 1,...,n)$. Put $$r = min(r_{1},...,r_{n})$$ and let $N$ be the neighborhood of $x$ of radius $r$. Then $N subset G_{i}$ for $i = 1,...,n$, so that $N subset H$, and $H$ is open.



$Box$



So I am having difficulty in seeing why we are able to say the neighborhood $N_{r}(x)$ is a subset of $G_{i}$ for every $i$. I am not following here. Any ideas?



Thanks.






analysis







share|cite|improve this question















Can someone help me through this theorem in 'Baby Rudin.'



For any collection $G_{1},...,G_{n}$ of open sets, $bigcaplimits_{i = 1}^{n} G_{i}$ is open.



Pf
Put $H = bigcaplimits_{i=1}^{n} G_{i}$. For any $x in H$, there exist neighborhoods $N_{i}$ of $x$, with radii $r_{i}$, such that $N_{i} subset G_{i} (i = 1,...,n)$. Put $$r = min(r_{1},...,r_{n})$$ and let $N$ be the neighborhood of $x$ of radius $r$. Then $N subset G_{i}$ for $i = 1,...,n$, so that $N subset H$, and $H$ is open.



$Box$



So I am having difficulty in seeing why we are able to say the neighborhood $N_{r}(x)$ is a subset of $G_{i}$ for every $i$. I am not following here. Any ideas?



Thanks.






analysis




analysis






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edited Nov 13 at 12:04









PiKindOfGuy

1299




1299










asked May 28 '14 at 14:03









zzz2991

1546




1546












  • The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
    – user122283
    May 28 '14 at 14:07










  • Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
    – user122283
    May 28 '14 at 14:10










  • Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
    – zzz2991
    May 28 '14 at 15:05


















  • The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
    – user122283
    May 28 '14 at 14:07










  • Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
    – user122283
    May 28 '14 at 14:10










  • Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
    – zzz2991
    May 28 '14 at 15:05
















The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
– user122283
May 28 '14 at 14:07




The "theorem" is actually trivial from the definition of a topology. I don't think you have learned that yet, though.
– user122283
May 28 '14 at 14:07












Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
– user122283
May 28 '14 at 14:10




Also, I don't have time to write a proof, but think intuitionistically, maybe in $mathbb{R}^2$.
– user122283
May 28 '14 at 14:10












Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
– zzz2991
May 28 '14 at 15:05




Thanks for the comment. That is where I am not following, geometrically. $r$ can be large no? I don't see what implies that $r$ is small enough.
– zzz2991
May 28 '14 at 15:05










1 Answer
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For each $i$, $r leq r_i$. Thus $N_r(x) subset N_i$ for each $i$. But $N_i subset G_i$ for each $i$. Thus it immediately follows that $N_r(x) subset G_i$ for each $i$.






share|cite|improve this answer





















  • Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
    – zzz2991
    May 28 '14 at 15:04










  • @zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
    – quid♦
    Aug 3 '16 at 20:42













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For each $i$, $r leq r_i$. Thus $N_r(x) subset N_i$ for each $i$. But $N_i subset G_i$ for each $i$. Thus it immediately follows that $N_r(x) subset G_i$ for each $i$.






share|cite|improve this answer





















  • Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
    – zzz2991
    May 28 '14 at 15:04










  • @zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
    – quid♦
    Aug 3 '16 at 20:42

















up vote
2
down vote













For each $i$, $r leq r_i$. Thus $N_r(x) subset N_i$ for each $i$. But $N_i subset G_i$ for each $i$. Thus it immediately follows that $N_r(x) subset G_i$ for each $i$.






share|cite|improve this answer





















  • Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
    – zzz2991
    May 28 '14 at 15:04










  • @zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
    – quid♦
    Aug 3 '16 at 20:42















up vote
2
down vote










up vote
2
down vote









For each $i$, $r leq r_i$. Thus $N_r(x) subset N_i$ for each $i$. But $N_i subset G_i$ for each $i$. Thus it immediately follows that $N_r(x) subset G_i$ for each $i$.






share|cite|improve this answer












For each $i$, $r leq r_i$. Thus $N_r(x) subset N_i$ for each $i$. But $N_i subset G_i$ for each $i$. Thus it immediately follows that $N_r(x) subset G_i$ for each $i$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 28 '14 at 14:26









Alex G.

6,115928




6,115928












  • Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
    – zzz2991
    May 28 '14 at 15:04










  • @zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
    – quid♦
    Aug 3 '16 at 20:42




















  • Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
    – zzz2991
    May 28 '14 at 15:04










  • @zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
    – quid♦
    Aug 3 '16 at 20:42


















Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
– zzz2991
May 28 '14 at 15:04




Thanks for the reply. Though, I don't see how we can say that $r$ is small enough so that the neighborhood of $N_{r}(x)$ is contained in the intersection. If that makes sense... I don't see how saying since $N_{r_{i}}(x)$ is a neighborhood in the respective $G_{i}$, we have that it is a neighborhood in the intersection. $r$ could be large, is what I am thinking...
– zzz2991
May 28 '14 at 15:04












@zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
– quid♦
Aug 3 '16 at 20:42






@zzz2991 you have $N_r(x) subset N_i subset G_i$ for every $i$. Thus $N_r(x) subset G_i$ for every $i$ thus $N_r(x)$ is in the intersection of the $G_i$.
– quid♦
Aug 3 '16 at 20:42




















 

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