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

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.

  • $\begingroup$ What is your question? $\endgroup$
    – Spencer
    Jun 22, 2014 at 21:55
  • $\begingroup$ The question was meant as: What is the proof? (See text parts in bold face.) $\endgroup$ Jun 22, 2014 at 22:08
  • $\begingroup$ Question still unclear? $\endgroup$ Jun 23, 2014 at 1:48
  • $\begingroup$ Let's suppose the Question is to show the equivalence of those two definitions of topological spaces. The verification can be broken down into a number of parts. You should attempt these, and ask about a specific point where you need assistance. $\endgroup$
    – hardmath
    Jun 23, 2014 at 2:35
  • $\begingroup$ No sorry, that vote was cast before your response. I've cast a reopen vote now that you have clarified it. $\endgroup$
    – Spencer
    Jun 23, 2014 at 2:35

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

  • $\begingroup$ @Downvoter: Could you please explain your concern! $\endgroup$ Jun 22, 2014 at 20:04
  • $\begingroup$ I did not downvote yet but: 1. Proofs written as long successions of implications are unreadable by humans without great efforts. 2. Already the first implication is wrong as written (what is $U_0$?). $\endgroup$
    – Did
    Jun 23, 2014 at 5:26
  • $\begingroup$ You're right - will think about changing my style for the future. Thx for the hint!! $\endgroup$ Jun 23, 2014 at 11:42
  • $\begingroup$ In my writing whenever a variable appears only on one side of an implication, then: no index means for all of these while an index zero means there exists this one. Is this what you meant or did you mean something else? $\endgroup$ Jun 23, 2014 at 11:52
  • 5
    $\begingroup$ A consequence of this idiosyncratic typographical convention is that your post is unreadable. $\endgroup$
    – Did
    Jun 23, 2014 at 12:20

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