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1. • CommentRowNumber1.
   • CommentAuthorporton
   • CommentTimeMar 11th 2014
   • (edited Mar 11th 2014)
   • PermaLink
   Author: porton
   Format: MarkdownItexIn <a href="http://ncatlab.org/nlab/show/proximity+space#uniform_spaces">this nWiki page</a> it's said: "Conversely, if $X$ is a proximity space, consider the family of sets of the form..." and then about finite families of sets. I feel if we would allow only pairs of set (finite families of cardinality $2$) instead of any finite families the resulted uniformity would not change. Is my conclusion that we can take only families of cardinality $2$ without change of the result correct? Or do I err? If it is correct, let's edit <a href="http://ncatlab.org/nlab/show/proximity+space#uniform_spaces">the page</a> to include this result. I would like to see the proof.

   In this nWiki page it’s said: “Conversely, if $X$ is a proximity space, consider the family of sets of the form…” and then about finite families of sets.

   I feel if we would allow only pairs of set (finite families of cardinality $2$) instead of any finite families the resulted uniformity would not change.

   Is my conclusion that we can take only families of cardinality $2$ without change of the result correct? Or do I err?

   If it is correct, let’s edit the page to include this result. I would like to see the proof.

2. • CommentRowNumber2.
   • CommentAuthorporton
   • CommentTimeMar 11th 2014
   • (edited Mar 11th 2014)
   • PermaLink
   Author: porton
   Format: MarkdownItexIt's enough to prove that $\bigcup_{i=1}^n ( A_i \times A_i)$ is representable as an intersection of sets of the form $( A^'_1 \times A^'_1) \cup ( A^'_2 \times A^'_2)$ (where $A^'_1$, $A^'_2$ conform to the formula for $A_1,\dots A_n$). Really, take $A^'_{1, i} = A_i$ and $A^'_{2, i} = \bigcup ( \{ A_1, \ldots A_n \} \setminus A_i)$. Then \[ \bigcup_{i=1}^n ( A_i \times A_i) = \bigcap \left\{ ( A^'_{1, i} \times A^'_{1, i}) \cup ( A^'_{2, i} \times A^'_{2, i}) \,|\, i=1,\dots n \right\} . \] It remains to prove the last formula.

   It’s enough to prove that $\bigcup_{i=1}^n ( A_i \times A_i)$ is representable as an intersection of sets of the form $( A^'_1 \times A^'_1) \cup ( A^'_2 \times A^'_2)$ (where $A^'_1$, $A^'_2$ conform to the formula for $A_1,\dots A_n$).

   Really, take $A^'_{1, i} = A_i$ and $A^'_{2, i} = \bigcup ( \{ A_1, \ldots A_n \} \setminus A_i)$. Then

   $$\bigcup_{i=1}^n ( A_i \times A_i) = \bigcap \left\{ ( A^'_{1, i} \times A^'_{1, i}) \cup ( A^'_{2, i} \times A^'_{2, i}) \,|\, i=1,\dots n \right\} .$$

   It remains to prove the last formula.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMar 12th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIf we used only pairs, then we would not have a [[base]] but only a sub-base.

   If we used only pairs, then we would not have a base but only a sub-base.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeMar 12th 2014
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexIf it is established that using only pairs gives a subbase that generates this base (or even generates the same uniformity as this base), then that would be worth remarking.

   If it is established that using only pairs gives a subbase that generates this base (or even generates the same uniformity as this base), then that would be worth remarking.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMar 12th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThe page [[base]] says that bases (and presumably subbases) for uniformities are trickier than topologies, but gives no details, and I don't know the details.

   The page base says that bases (and presumably subbases) for uniformities are trickier than topologies, but gives no details, and I don’t know the details.

6. • CommentRowNumber6.
   • CommentAuthorporton
   • CommentTimeMar 13th 2014
   • PermaLink
   Author: porton
   Format: MarkdownItexIt's false: See <a href="http://math.stackexchange.com/questions/711100/replace-n-sets-with-two-sets-set-theoretic-equality/711151">this math.SE question</a>.

   It’s false: See this math.SE question.

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeMar 19th 2014
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexBases and subbases for uniformities (given by entourages) are described at [[uniform space]]. The tricky bit is that axioms 0--3 (as numbered there) are already a bit oversaturated (for simplicity), so you have to weaken these to get a proper notion of subbase. (Bases themselves are not tricky.)

   Bases and subbases for uniformities (given by entourages) are described at uniform space. The tricky bit is that axioms 0–3 (as numbered there) are already a bit oversaturated (for simplicity), so you have to weaken these to get a proper notion of subbase. (Bases themselves are not tricky.)

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeMar 19th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexI added a note to the page [[base]] that bases and subbases for uniformities are described at [[uniform space]].

   I added a note to the page base that bases and subbases for uniformities are described at uniform space.