📚 Recommended Mathematics Books
Topology (Munkres) | General Topology (Engelking) | Counterexamples in Topology | Rudin's AnalysisAs an Amazon associate, I earn from qualifying purchases.
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.
🔬 Advanced Mathematics References
- Sheaves in Geometry and Logic
- Categories for the Working Mathematician
- Stone Spaces
- Algebraic Topology (Hatcher)
- Concrete Mathematics
As an Amazon associate I refer fromm qualifying purchases.