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How to prove the product of totally bounded uniform spaces is totally bounded?
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One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;
Lemma 1. the product of countable totally bounded metric spaces is totally bounded.
Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.
Lemma 3. The product of uniform spaces is uniform induced.
How about the case when uncountable?
general-topology uniform-spaces
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
asked Jul 18 at 8:06
Shen Chong
386
add a comment |Â
up vote
2
down vote
favorite
One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;
Lemma 1. the product of countable totally bounded metric spaces is totally bounded.
Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.
Lemma 3. The product of uniform spaces is uniform induced.
How about the case when uncountable?
general-topology uniform-spaces
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
asked Jul 18 at 8:06
Shen Chong
386
1
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;
Lemma 1. the product of countable totally bounded metric spaces is totally bounded.
Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.
Lemma 3. The product of uniform spaces is uniform induced.
How about the case when uncountable?
general-topology uniform-spaces
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
asked Jul 18 at 8:06
Shen Chong
386
One should note that the family here may not countable. If it is countable, the it is a consequence of the following results;
Lemma 1. the product of countable totally bounded metric spaces is totally bounded.
Lemma 2. Every uniform space is uniformly isomorphic to a product of metrizable uniform spaces.
Lemma 3. The product of uniform spaces is uniform induced.
How about the case when uncountable?
general-topology uniform-spaces
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
asked Jul 18 at 8:06
Shen Chong
386
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
edited Jul 18 at 21:13
Henno Brandsma
91.6k342100
91.6k342100
asked Jul 18 at 8:06
Shen Chong
386
asked Jul 18 at 8:06
Shen Chong
386
asked Jul 18 at 8:06
Shen Chong
386
386
1
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28
add a comment |Â
1
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28
1
1
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
down vote
Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, mathcalU_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=prod_s in S X_s$ which is of the form $U=(x_s,y_s)_s in S: $, where $s_1,ldots, s_N$ are finitely many indices from $S$ and $V_s_i in mathcalU_s_i$. For each $iin 1,ldots,N$ we can find a finite set $F_i subseteq X_s_i$ that forms is $V_s_i$-dense. Fix any point $p in X$ for "defaultness" and define $$F = (x_s) in S: forall i in 1,ldots,N: x_s_i in F_i text and forall s in Ssetminuss_1,ldots,s_N: x_s =p_s$$
and note that $|F| = prod_i=1^N |F_i|$ and hence is finite and is the required finite $U$-dense set.
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
add a comment |Â
up vote
0
down vote
The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = Pi_s in S X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = Pi_s in S X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
answered Jul 18 at 9:08
Paul Frost
3,703420
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, mathcalU_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=prod_s in S X_s$ which is of the form $U=(x_s,y_s)_s in S: $, where $s_1,ldots, s_N$ are finitely many indices from $S$ and $V_s_i in mathcalU_s_i$. For each $iin 1,ldots,N$ we can find a finite set $F_i subseteq X_s_i$ that forms is $V_s_i$-dense. Fix any point $p in X$ for "defaultness" and define $$F = (x_s) in S: forall i in 1,ldots,N: x_s_i in F_i text and forall s in Ssetminuss_1,ldots,s_N: x_s =p_s$$
and note that $|F| = prod_i=1^N |F_i|$ and hence is finite and is the required finite $U$-dense set.
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, mathcalU_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=prod_s in S X_s$ which is of the form $U=(x_s,y_s)_s in S: $, where $s_1,ldots, s_N$ are finitely many indices from $S$ and $V_s_i in mathcalU_s_i$. For each $iin 1,ldots,N$ we can find a finite set $F_i subseteq X_s_i$ that forms is $V_s_i$-dense. Fix any point $p in X$ for "defaultness" and define $$F = (x_s) in S: forall i in 1,ldots,N: x_s_i in F_i text and forall s in Ssetminuss_1,ldots,s_N: x_s =p_s$$
and note that $|F| = prod_i=1^N |F_i|$ and hence is finite and is the required finite $U$-dense set.
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, mathcalU_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=prod_s in S X_s$ which is of the form $U=(x_s,y_s)_s in S: $, where $s_1,ldots, s_N$ are finitely many indices from $S$ and $V_s_i in mathcalU_s_i$. For each $iin 1,ldots,N$ we can find a finite set $F_i subseteq X_s_i$ that forms is $V_s_i$-dense. Fix any point $p in X$ for "defaultness" and define $$F = (x_s) in S: forall i in 1,ldots,N: x_s_i in F_i text and forall s in Ssetminuss_1,ldots,s_N: x_s =p_s$$
and note that $|F| = prod_i=1^N |F_i|$ and hence is finite and is the required finite $U$-dense set.
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
Every factor space embeds into the product as a subspace, so the previous theorem in Engelking's General Topology (which you're quoting here) implies the left to right implication right away.
To check the right to left one, let all $(X_s, mathcalU_s)$ be totally bounded. It suffices to check the definition of total boundedness for a basic entourage for $X=prod_s in S X_s$ which is of the form $U=(x_s,y_s)_s in S: $, where $s_1,ldots, s_N$ are finitely many indices from $S$ and $V_s_i in mathcalU_s_i$. For each $iin 1,ldots,N$ we can find a finite set $F_i subseteq X_s_i$ that forms is $V_s_i$-dense. Fix any point $p in X$ for "defaultness" and define $$F = (x_s) in S: forall i in 1,ldots,N: x_s_i in F_i text and forall s in Ssetminuss_1,ldots,s_N: x_s =p_s$$
and note that $|F| = prod_i=1^N |F_i|$ and hence is finite and is the required finite $U$-dense set.
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
answered Jul 18 at 21:27
Henno Brandsma
91.6k342100
91.6k342100
add a comment |Â
add a comment |Â
up vote
0
down vote
The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = Pi_s in S X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = Pi_s in S X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
answered Jul 18 at 9:08
Paul Frost
3,703420
add a comment |Â
up vote
0
down vote
The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = Pi_s in S X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = Pi_s in S X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
answered Jul 18 at 9:08
Paul Frost
3,703420
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = Pi_s in S X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = Pi_s in S X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
answered Jul 18 at 9:08
Paul Frost
3,703420
The proof depends on which theorems you allow to use. You certainly can give a direct proof, but in my opinion the most elegant way is this:
Theorems used:
(1) Each uniform space $X$ has a completion $X'$ (i.e. a complete uniform space containing $X$ as a dense subset).
(2) If $X$ is totally bounded, then $X'$ is totally bounded.
(3) A uniform space is compact if and only if it is complete and totally bounded.
Proof:
If $P = Pi_s in S X_s$ is totally bounded, then each $X_s$ is totally bounded because it embeds as a uniform subspace into $P$.
If all $X_s$ are totally bounded, then the completions $X'_s$ are compact. Therefore $P' = Pi_s in S X'_s$ is compact. Since $P$ embeds as a uniform subspace of $P'$, it is totally bounded.
answered Jul 18 at 9:08
Paul Frost
3,703420
edited Jul 18 at 22:41
answered Jul 18 at 9:08
Paul Frost
3,703420
answered Jul 18 at 9:08
Paul Frost
3,703420
answered Jul 18 at 9:08
Paul Frost
3,703420
3,703420
add a comment |Â
add a comment |Â
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1
Have you tried the naive approach of simply taking one entourage in the product, seeing what it looks like in terms of entourages of the spaces, use their total boundedness, and then go back up ?
– Max
Jul 18 at 8:45
The countable case has nothing to do with metrisable spaces at all.The proof is the same regardless of index set.
– Henno Brandsma
Jul 18 at 21:12
The sketch of the proof is above 8.3.2. and 8.3.3. As usual, the simple details are left as an exercise.
– Henno Brandsma
Jul 18 at 21:28