Topology: Opens vs Neighborhoods

Question

Disclaimer: This thread is meant informative and therefore written in Q&A style. The problems are highlighted in bold face.

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The axiomatization of topology can be done in various ways all of them having their own advantage. Here I would like to investigate two of them specifically.

There's the one by open sets usually given: $$\bullet \#I<\infty:\quad A_i\in\mathcal{T}\implies \bigcap_{i\in I}A\in\mathcal{T}\\ \bullet \#I\leq\infty:\quad A_i\in\mathcal{T}\implies\bigcup_{i\in I}A_i\in\mathcal{T}$$ and the one by neighborhoods introduced by Felix Hausdorff: $$\bullet A\subseteq B:\quad A\in\mathcal{N}(x)\implies B\in\mathcal{N}(x)\\ \bullet A,B\in\mathcal{N}(x)\implies A\cap B\in\mathcal{N}(x)\\ \bullet \forall x\in X:\quad\mathcal{N}(x)\neq\{\}\\ \bullet A\in\mathcal{N}(x)\implies x\in A\\ \bullet A\in\mathcal{N}(x)\implies\exists C_0\in\mathcal{N}:\quad A\in\mathcal{N}(c)\text{ for all }c\in C_0(x)$$ Prove that any family of open sets give rise to a neighborhood system via: $$A\in\mathcal{N_T}(x):\iff\exists U_0\in\mathcal{T}:\quad x\in U_0\subseteq A\quad$$ and that any neighborhood system gives rise to a family of open sets via: $$A\in\mathcal{T_N}:\iff\forall a\in A:\quad A\in\mathcal{N}(a)$$ Moreover prove that their equivalent in the sense: $$\mathcal{T}\mapsto\mathcal{N_T}\mapsto\mathcal{T}\text{ and }\mathcal{N}\mapsto\mathcal{T_N}\mapsto\mathcal{N}$$ (Note that both must be checked in order to ensure injectivity and surjectivity.)

So we can switch back and forth between both descriptions for topology. Here are two situations where this is exploited:

a. The interior is defined via neighborhoods: $$A^\circ:=\{z:A\in\mathcal{N}(z)\}$$ It is contained and open (see Topology: Interior): $$A^\circ\subseteq A\text{ and }A^\circ\in\mathcal{N}(z)\text{ for all }z\in A^\circ$$ Therefore neighborhoods have nonempty interior: $$A^\circ=\bigcup_{A\supseteq U\in\mathcal{T}}U$$ b. Continuity is defined via neighborhoods: $$N\in\mathcal{N}(f(x))\implies f^{-1}N\in\mathcal{N}(x)$$ Thus in locally convex spaces topology is entailed fully in any point: $$N\in\mathcal{N}(x)\iff N+a\in\mathcal{N}(x+a)$$

So while open sets reflect general aspects of topology correlations between space itself and topology become lucid via neighborhoods.

Answer 1

For better reading I left out the details...

Any family of open sets gives rise to a neighborhood system since: $$\bullet\left(A\in\mathcal{N_T}(x)\right)\implies\left(U_0\subseteq A\quad x\in U_0\in\mathcal{T}\right)\\\implies\left(U_0\subseteq B\quad x\in U_0\in\mathcal{T}\right)\implies\left(B\in\mathcal{N_T}(x)\right)\quad A\subseteq B\\ \bullet\left(A,B\in\mathcal{N_T}(x)\right)\implies\left(U_A\subseteq A,U_B\subseteq B\quad x\in U_A,U_B\in\mathcal{T}\right)\\\implies\left(U_A\cap U_B\subseteq A\cap B\quad x\in U_A\cap U_B\in\mathcal{T}\right)\implies\left(A\cap B\in\mathcal{N_T}(x)\right)\\ \bullet\left(X\in\mathcal{T}:\quad x\in X\subseteq X\right)\implies\left(X\in\mathcal{N_T}(x)\right)\\ \bullet\left(A\in\mathcal{N_T}(x)\right)\implies\left(U_0\subseteq A\quad x\in U_0\in\mathcal{T}\right)\implies\left(x\in U_0\subseteq A\right)\\ \bullet\left(A\in\mathcal{N_T}(x)\right)\implies\left(U_0\subseteq A\quad x\in U_0\in\mathcal{T}\right)\\\implies\left(U_0\subseteq A,U_0\subseteq U_0\quad u,x\in U_0\in\mathcal{T}\right)\implies\left(A\in\mathcal{N}(u)\quad u\in U_0\in\mathcal{N_T}(x)\right)$$

Any family of open sets gives rise to a neighborhood system since: $$\left(A,B\in\mathcal{T_N}\right)\implies\left(A\in\mathcal{N}(a),B\in\mathcal{N}(b)\quad a\in A,b\in B\right)\\\implies\left(A\cap B\in\mathcal{N}(c)\quad c\in A\cap B\right)\implies\left(A\cap B\in\mathcal{T_N}\right)\\ \left(A_i\in\mathcal{T_N}\right)\implies\left(A_i\in\mathcal{N}(a_i)\quad a_i\in A_i\right)\\\implies\left(\bigcup_{i\in I}A_i\in\mathcal{N}(c)\quad c\in\bigcup_{i\in I}A_i\right)\implies\left(\bigcup_{i\in I}A_i\in\mathcal{T_N}\right)\\ \left(\varnothing\in\mathcal{N}(x)\quad x\in\varnothing\right)\implies\left(\varnothing\in\mathcal{T_N}\right)\\ \left(\mathcal{N}(x)\neq\{\}\quad x\in X\right)\implies\left(X\in\mathcal{N}(x)\quad x\in X\right)\implies\left(X\in\mathcal{T_N}\right)$$

Not that the interior is contained and open: $$A^\circ\subseteq A\text{ and }A^\circ\in\mathcal{N}(z)\quad z\in A^\circ$$ (Its precise definition, statement and proof can be found in Topology: Interior.)

Moreover their equivalent since: $$\left(A\in\mathcal{N_T}(x)\right)\implies\left(U_0\subseteq A\quad x\in U_0\in\mathcal{T_N}\right)\\\implies\left(U_0\subseteq A\quad x\in U_0\in\mathcal{N}(u)\quad u\in U_0\right)\implies\left(U_0\subseteq A\quad U_0\in\mathcal{N}(x)\right)\implies\left(A\in\mathcal{N}(x)\right)\\ \left(A\in\mathcal{N}(x)\right)\implies\left(A^\circ\subseteq A\quad x\in A^\circ\in\mathcal{N}(z)\quad z\in A^\circ\right)\\\implies\left(A^\circ\subseteq A\quad x\in A^\circ\in\mathcal{T_N}\right)\implies\left(A\in\mathcal{N_T}(x)\right)$$ $$\left(A\in\mathcal{T_N}\right)\implies\left(A\in\mathcal{N_T}(a)\quad a\in A\right)\\\implies\left(U_a\subseteq A\quad a\in U_a\in\mathcal{T}\quad a\in A\right)\implies\left(A=\bigcup_{a\in A}U_a\in\mathcal{T}\right)\\ \left(A\in\mathcal{T}\right)\implies\left(A\subseteq A\quad a\in A\in\mathcal{T}\quad a\in A\right)\implies\left(A\in\mathcal{N_T}(a)\quad a\in A\right)\implies\left(A\in\mathcal{T_N}\right)$$