📚 Recommended Mathematics Books

Topology (Munkres)  |  General Topology (Engelking)  |  Counterexamples in Topology  |  Rudin's Analysis

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I have a little generalized the following old theorem:

$latex (a\sqcap^{\mathfrak{A}}b)^{\ast}=(a\sqcap^{\mathfrak{A}}b)^{+}=a^{\ast}\sqcup^{\mathfrak{A}}b^{\ast}=a^{+}\sqcup^{\mathfrak{A}}b^{+}$.

I have also found a new (easy to prove) theorem:

$latex (a\sqcup^{\mathfrak{A}}b)^{\ast}=(a\sqcup^{\mathfrak{A}}b)^{+}=a^{\ast}\sqcap^{\mathfrak{A}}b^{\ast}=a^{+}\sqcap^{\mathfrak{A}}b^{+}$.

The above formulas hold for filters on a set (and some generalizations).

Do these formulas hold also for funcoids? (an interesting conjecture)

See my free e-book.