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
edited Nov 13 at 12:04
PiKindOfGuy
1299
asked May 28 '14 at 14:03
zzz2991
1546
add a comment |
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
edited Nov 13 at 12:04
PiKindOfGuy
1299
asked May 28 '14 at 14:03
zzz2991
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
add a comment |
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
edited Nov 13 at 12:04
PiKindOfGuy
1299
asked May 28 '14 at 14:03
zzz2991
1546
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
edited Nov 13 at 12:04
PiKindOfGuy
1299
asked May 28 '14 at 14:03
zzz2991
1546
edited Nov 13 at 12:04
PiKindOfGuy
1299
asked May 28 '14 at 14:03
zzz2991
1546
edited Nov 13 at 12:04
PiKindOfGuy
1299
edited Nov 13 at 12:04
PiKindOfGuy
1299
edited Nov 13 at 12:04
PiKindOfGuy
1299
1299
asked May 28 '14 at 14:03
zzz2991
1546
asked May 28 '14 at 14:03
zzz2991
1546
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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 $i$.
answered May 28 '14 at 14:26
Alex G.
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $i$.
answered May 28 '14 at 14:26
Alex G.
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
add a comment |
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 $i$.
answered May 28 '14 at 14:26
Alex G.
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
add a comment |
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 $i$.
answered May 28 '14 at 14:26
Alex G.
6,115928
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 $i$.
answered May 28 '14 at 14:26
Alex G.
6,115928
answered May 28 '14 at 14:26
Alex G.
6,115928
answered May 28 '14 at 14:26
Alex G.
6,115928
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
add a comment |
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
add a comment |
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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