ω 1 -strongly compact cardinals and cardinal functions in topology Toshimichi Usuba Waseda University Nov. 19, 2018 Reflections on Set Theoretic reflection, Sant Bernat T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 1 / 27
Reflection and Compactness Reflection: LARGE to SMALL, GLOBAL to LOCAL Compactness: SMALL to LARGE, LOCAL to GLOBAL These are dual concepts! T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 2 / 27
ω 1 -strongly compact cardinal Definition An uncountable cardinal κ is strongly compact if for every κ -complete filter F over the set A there is a κ -complete ultrafilter over A extending F . Strongly compact cardinals give natural limitations or upper bounds in many contexts. Bagaria and Magidor introduced the notion of ω 1 -strongly compact cardinal to analyze and optimize it. Definition (Bagaria-Magidor (2014)) An uncountable cardinal κ is ω 1 -strongly compact if for every κ -complete filter F over the set A there is a σ -complete ultrafilter over A extending F . T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 3 / 27
ω 1 -strongly compact cardinal Definition An uncountable cardinal κ is strongly compact if for every κ -complete filter F over the set A there is a κ -complete ultrafilter over A extending F . Strongly compact cardinals give natural limitations or upper bounds in many contexts. Bagaria and Magidor introduced the notion of ω 1 -strongly compact cardinal to analyze and optimize it. Definition (Bagaria-Magidor (2014)) An uncountable cardinal κ is ω 1 -strongly compact if for every κ -complete filter F over the set A there is a σ -complete ultrafilter over A extending F . T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 3 / 27
Fact (Bagaria-Magidor) 1 If κ is ω 1 -strongly compact, then every cardinal ≥ κ is ω 1 -strongly compact. 2 So the important one is the least ω 1 -strongly compact cardinal. 3 strongly compact ⇒ ω 1 -strongly compact. 4 If κ is ω 1 -strongly compact, then there is a measurable cardinal ≤ κ . Fact (Bagaria-Magidor) 1 It is consistent that the least ω 1 -strongly compact is supercompact. 2 It is consistent that the least measurable is ω 1 -strongly compact (in this case, the least measurable must be strongly compact). 3 It is consistent that the least ω 1 -strongly compact is singular. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 4 / 27
Fact (Bagaria-Magidor) 1 If κ is ω 1 -strongly compact, then every cardinal ≥ κ is ω 1 -strongly compact. 2 So the important one is the least ω 1 -strongly compact cardinal. 3 strongly compact ⇒ ω 1 -strongly compact. 4 If κ is ω 1 -strongly compact, then there is a measurable cardinal ≤ κ . Fact (Bagaria-Magidor) 1 It is consistent that the least ω 1 -strongly compact is supercompact. 2 It is consistent that the least measurable is ω 1 -strongly compact (in this case, the least measurable must be strongly compact). 3 It is consistent that the least ω 1 -strongly compact is singular. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 4 / 27
Lindel¨ of degree Definition X : topological space The Lindel¨ of degree of X , L ( X ), is min { κ | every open cover of X has a subcover of size ≤ κ } . X is Lindel¨ of if L ( X ) = ω , that is, every open cover has a countable subcover. Clearly | X | ≥ L ( X ). T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 5 / 27
Product of Lindel¨ of spaces Fact 1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨ of spaces need not to be Lindel¨ of; If S is the of but L ( S 2 ) = 2 ω . Sorgenfrey line, then S is Lindel¨ Question (Classical question) How large is the Lindel¨ of degree of the product of Lindel¨ of spaces? Fact (Juh´ asz(?)) For Lindel¨ of spaces X i ( i ∈ I ), L ( ∏ i ∈ I X i ) ≤ the least strongly compact cardinal. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 6 / 27
Product of Lindel¨ of spaces Fact 1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨ of spaces need not to be Lindel¨ of; If S is the of but L ( S 2 ) = 2 ω . Sorgenfrey line, then S is Lindel¨ Question (Classical question) How large is the Lindel¨ of degree of the product of Lindel¨ of spaces? Fact (Juh´ asz(?)) For Lindel¨ of spaces X i ( i ∈ I ), L ( ∏ i ∈ I X i ) ≤ the least strongly compact cardinal. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 6 / 27
Product of Lindel¨ of spaces Fact 1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨ of spaces need not to be Lindel¨ of; If S is the of but L ( S 2 ) = 2 ω . Sorgenfrey line, then S is Lindel¨ Question (Classical question) How large is the Lindel¨ of degree of the product of Lindel¨ of spaces? Fact (Juh´ asz(?)) For Lindel¨ of spaces X i ( i ∈ I ), L ( ∏ i ∈ I X i ) ≤ the least strongly compact cardinal. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 6 / 27
Product of Lindel¨ of spaces Fact 1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨ of spaces need not to be Lindel¨ of; If S is the of but L ( S 2 ) = 2 ω . Sorgenfrey line, then S is Lindel¨ Question (Classical question) How large is the Lindel¨ of degree of the product of Lindel¨ of spaces? Fact (Juh´ asz(?)) For Lindel¨ of spaces X i ( i ∈ I ), L ( ∏ i ∈ I X i ) ≤ the least strongly compact cardinal. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 6 / 27
Characterization: Product of Lindel¨ of spaces Fact (Bagaria-Magidor) For uncountable cardinal κ , the following are equivalent: 1 κ is ω 1 -strongly compact. 2 For every family { X i | i ∈ I } of Lindel¨ of spaces and open cover U of the product space ∏ i ∈ I X i , U has a subcover of size < κ . 3 For every family { X i | i ∈ I } of Lindel¨ of spaces we have L ( ∏ i ∈ I X i ) ≤ κ , the least ω 1 -strongly compact= sup { L ( ∏ i ∈ I X i ) | X i is Lindel¨ of } . (1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω 1 -strongly compact cardinal, then we can find κ ≥ λ with L ( ω κ ) ≥ λ (use infinite Abelian group theory). T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 7 / 27
Characterization: Product of Lindel¨ of spaces Fact (Bagaria-Magidor) For uncountable cardinal κ , the following are equivalent: 1 κ is ω 1 -strongly compact. 2 For every family { X i | i ∈ I } of Lindel¨ of spaces and open cover U of the product space ∏ i ∈ I X i , U has a subcover of size < κ . 3 For every family { X i | i ∈ I } of Lindel¨ of spaces we have L ( ∏ i ∈ I X i ) ≤ κ , the least ω 1 -strongly compact= sup { L ( ∏ i ∈ I X i ) | X i is Lindel¨ of } . (1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω 1 -strongly compact cardinal, then we can find κ ≥ λ with L ( ω κ ) ≥ λ (use infinite Abelian group theory). T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 7 / 27
Characterization: Product of Lindel¨ of spaces Fact (Bagaria-Magidor) For uncountable cardinal κ , the following are equivalent: 1 κ is ω 1 -strongly compact. 2 For every family { X i | i ∈ I } of Lindel¨ of spaces and open cover U of the product space ∏ i ∈ I X i , U has a subcover of size < κ . 3 For every family { X i | i ∈ I } of Lindel¨ of spaces we have L ( ∏ i ∈ I X i ) ≤ κ , the least ω 1 -strongly compact= sup { L ( ∏ i ∈ I X i ) | X i is Lindel¨ of } . (1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω 1 -strongly compact cardinal, then we can find κ ≥ λ with L ( ω κ ) ≥ λ (use infinite Abelian group theory). T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 7 / 27
Bagaria and Magidor obtained various characterizations of ω 1 -strongly compact cardinal in many fields: Infinite Abelian groups. Topological spaces. Reflection principles... In this talk we consider more characterizations of ω 1 -strongly compact cardinals. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 8 / 27
Characterization:Infinitary logic Fact (Folklore) κ is ω 1 -strongly compact ⇐ ⇒ κ is L ω 1 ,ω -compact; that is, for every theory T of L ω 1 ω -sentences, T has a model if every subtheory of T with size < κ has a model. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 9 / 27
Characterization:Uniform filters A filter over the infinite cardinal κ is uniform if | A | = κ for every A ∈ F . Fact (Ketonen) An uncountable cardinal κ is strongly compact if and only if for every regular λ ≥ κ , there is a κ -complete uniform ultrafilter over λ . Proposition An uncountable cardinal κ is ω 1 -strongly compact if and only if for every regular λ ≥ κ , there is a σ -complete uniform ultrafilter over λ . Proof is the same to Ketonen’s one. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 10 / 27
Characterization:Uniform filters A filter over the infinite cardinal κ is uniform if | A | = κ for every A ∈ F . Fact (Ketonen) An uncountable cardinal κ is strongly compact if and only if for every regular λ ≥ κ , there is a κ -complete uniform ultrafilter over λ . Proposition An uncountable cardinal κ is ω 1 -strongly compact if and only if for every regular λ ≥ κ , there is a σ -complete uniform ultrafilter over λ . Proof is the same to Ketonen’s one. T. Usuba (Waseda Univ.) ω 1 -strongly compact Nov.19, 2018 10 / 27
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