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The German Wikipedia page on topological boundary [1] states that the boundary operator $\partial$ can be characterized by the following four axioms:

  1. $\partial\emptyset = \emptyset$,
  2. $\partial(\partial U) \subseteq \partial U$,
  3. $\partial(U^\complement)= \partial U$, and
  4. $(U\cap V)\cap \partial(U\cap V) = (U\cap V)\cap (\partial U\cup \partial V)$

for each $U,V\subseteq X$

Is this really correct? There are two references given of which one is available online [2]. But the author apparently distinguishes between boundary and frontier. The author defines the interior $\mathbf{int}(S)$ of a set $S$ as the largest open subset of $S$; the boundary is defined as "[the union of] the points of $S$ which are not interior points constitute [the boundary] $\mathbf{bd}(S)$." The author proceeds and defines the frontier $\mathbf{fr}(S)$ of $S$ as "the union of $\mathbf{bd}(S)$ and $\mathbf{bd}(S')$", where $S'$ denotes the complement of $S$ in $X$. (Note that I replaced $X$ with $S$ to be constitent with the above quote from Wikipedia).

I understand the definitions as \begin{align*} \mathbf{bd}(S) &= S\setminus\mathbf{int}(S) \qquad\text{and}\\ \mathbf{fr}(S) &= \mathbf{bd}(S)\cup \mathbf{bd}(S^\complement).\end{align*} Consider $\mathbb R$ together with the usual topology, and let $S = [0, 1)$. Then $\mathbf{bd}([0,1)) = \{0\}$, and $\mathbf{fr}([0,1)) = \{0,1\}$. I see the boundary $\partial S$ of the set $S$ typically defined as $\partial S = \mathbf{cls}(S) \cap \mathbf{cls}(S^\complement)$, or, equivalently, defined as $\partial S = \mathbf{cls}(S) \cap \mathbf{int}(S)^\complement$. Since $\partial([0,1)) = \{0,1\}$, I suppose that $\partial = \mathbf{fr}$. Is this correct? I am skeptical because I usually see the boundary operator $\partial$ axiomatized by

  1. $\partial \emptyset = \emptyset$,
  2. $\partial(\partial U) \subseteq \partial U$,
  3. $\partial(U^\complement) = \partial U$,
    1. $\partial U\subseteq V\cup \partial V$ if $U\subseteq V$, and
    2. $\partial (U\cup V) \subseteq \partial U\cup \partial V$

for all $U,V\subseteq X$. But I don't see how to derive 4. from 4.1. and 4.2., as well as the converse, i.e. derive 4.1 and 4.2 from 4.

So, do both axiomatizations agree with each other? And if yes, how can I show it? This is far from obvious for me.

References

[1] https://de.wikipedia.org/wiki/Rand_(Topologie)#Randaxiome

[2] Vaidyanathaswamy: Set topology. 1964, p. 57–58. URL: https://books.google.de/books?id=yDMipybQ64kC&printsec=frontcover&hl=de#v=onepage&q&f=false

1 Answers1

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This posting is not a complete answer. It only discusses the first axiomatization in the OP.

The first axiomatization correspond to the usual notion of boundary defined as $\overline{A}\cap\overline{X\setminus A}$. To see this, suppose $(X,\tau)$ is a topological space, that is $X$ is a non empty set and $\tau$ a collection of subsets of a set $X$ such that

  1. for any $\mathcal{U}\subset \tau$, $\bigcup \mathcal{U}\in \tau$,
  2. for any finite collection $\mathcal{J}\subset\tau$ $\bigcap\mathcal{J}\in \tau$ (if $\mathcal{J}=\emptyset$, it is understood that $\bigcap\mathcal{J}=X$).

Define $F$ to be a closed set iff $X\setminus F\in \tau$; for any set $A\subset X$, the closure $\overline{A}$ of $A$ is the intersection of all closed subsets of $X$ that contain $A$; the interior $A^o$ of $A$ is the union of all sets in $\tau$ that are contained in $A$; the boundary $\partial A$ of $A$ is defined as $\overline{A}\cap\overline{X\setminus A}$.

It follows easily that $(X\setminus\overline{A})=(X\setminus A)^o$ for all $A\subset X$ and from this,

  • $\partial A=\partial(X\setminus A)=\overline{A}\setminus A^o$;
  • $\partial(A\cup B)\subset \big(\partial A\big)\cup\big(\partial B\big)$;
  • $\partial (A\cap B)\subset \big(\partial A\big)\cup\big(\partial B\big)$.

These properties of $\partial$ in turn imply that

  • $\overline{A}=A^o\cup\partial{A}\subset A\cup\partial{A}\subset\overline{A}$.
  • $\partial(\overline{A})=\overline{\overline{A}}\cap\overline{X\setminus\overline{A}}= \overline{A}\cap\overline{(X\setminus A)^o}\subset\overline{A}\cap\overline{X\setminus A}=\partial A$
  • $\partial(\partial A)=\partial(\overline{A}\cap\overline{X\setminus A})\subset \partial{A}\cup\partial(X\setminus A)=\partial A$
  • $(A\cap B)\cap\partial(A\cap B)=A\cap B\cap(\overline{A\cap B}\cap\overline{(X\setminus A)\cup(X\setminus B)}=(A\cap B)\cap(\overline{X\setminus A}\cup\overline{X\setminus B})=(A\cap B\cap\overline{X\setminus A}) \cup (A\cap B\cap\overline{X\setminus B})=(A\cap B)\cap(\partial{A}\cup\partial{B})$

In summary,

Proposition 1: Given a topological space $(X,\tau)$, the boundary operator $\partial: \mathcal{P}(X)\rightarrow\mathcal{P}(X)$ defined as $\partial A =\overline{A}\cap\overline{X\setminus A}$ satisfies

  1. $\partial\emptyset=\emptyset$
  2. $\partial{A}=\partial(X\setminus A)$
  3. $\partial(\partial A)\subset \partial A$
  4. $(A\cap B)\cap\partial(A\cap B)=(A\cap B)\cap(\partial A\cup\partial B)$

Notice that a set $A$ is open ($A=A^o$) iff $A\cap \partial{A}=\emptyset$. Indeed $A\cap\partial{A}=A\setminus A^o$.

Conversely,

Proposition 2: Given a nonempty set $X$, suppose $\beta:\mathcal{P}(X)\rightarrow\mathcal{P}(X)$ satisfies properties (1) through (4) in Proposition 2, then $\mathscr{G}:=\{G\subset X: G\cap \beta(G)=\emptyset\}$ is topology on $X$, and in such topology, $\partial A = \beta(A)$ for all $A\subset X$.

This axiomatization based in the boundary operator is mentioned in for example Dugundji, J. Topology, William C Brown Pub, 1966, chapter 3 section 5. The details (proof of Proposition 2) are left there to the reader.

A detailed analysis of this axiomatization and a proof of Proposition 2 can be found in Wang, M.L., Relations among basic concepts in topology, M.S. Thesis, Oregon State U., 1968, pp. 24-28.


Here are other related concepts related to the boundary $\partial A$ of a set $A$.

  • In many texts (especially in the French school of topology), $\partial A=\overline{A}\cap\overline{X\setminus A}$ is called frontier and is denoted by $\operatorname{Fr}(A)$.
  • The set $\partial_I A:=A\cap\partial{A}=A\setminus(\overline{X\setminus A})=A\setminus A^o$ is called the internal boundary
  • The set $\partial_EA:=\overline{A}\setminus A=(X\setminus A)\cap\partial (X\setminus A)=\partial_I(X\setminus A)$ is called the external boundary of $A$.
  • Notice that $$\partial_IA\cup\partial_EA=(A\cap\overline{X\setminus A})\cup((X\setminus A)\cap \overline{A})=\partial A$$

The axiomatization described in reference [2] in the OP is related to internal boundary operator.


That the first axiomatization implies the second is now trivial by virtue of Proposition 2, for if $U\subset V$, then $$\partial U\subset \overline{U}=U\cup\partial U\subset \overline{V}=V\cup\partial{V}$$

It is known that the second axiomatization produces a topology $\mathscr{G}'$ on $X$ for which $\partial{A}=\overline{A}\cap{X\setminus A}$, where the closure is with respect to $\mathscr{G}'$. In particular $A\mapsto\partial A$ satisfies the axioms in the first axiomatization.

Thus, indeed, both axiomatizations are equivalent.

Mittens
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  • How did you get from $A\cap B\cap (\overline{A\cap B} \cap \overline{(X\setminus A)\cup (X\setminus B)})$ to $A\cap B\cap (\overline{(X\setminus A)} \cup \overline{(X\setminus B)})$ ? What happened to $\overline{A\cap B}$ ? – Syd Amerikaner Mar 08 '22 at 15:12
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    @SydAmerikaner: Notice that $(A\cap B)\cap\overline{A\cap B}=A\cap B$; and that $\overline{P\cup Q}=\overline{P}\cup\overline{Q}$ for any sets $P,Q\subset X$. – Mittens Mar 08 '22 at 18:34
  • I just noted your edit. This is pretty much I was looking for. Thank you very much for your effort! – Syd Amerikaner Mar 09 '22 at 18:08
  • You point out that the second axiomatization (consisting of 1., 2., 3., 4.1, and 4.2) is related to the internal boundary operator. These boundary operators all give raise to the same topology. However, the operators are not equivalent in the sense that I can replace 4. in the first axiomatization with 4.1 and 4.2 from the second. Is this correct? – Syd Amerikaner Mar 09 '22 at 22:49
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    @SydAmerikaner: I was referring actually to the axiomatization in reference [2] of your OP where the author defines $bd(A)=A\cap\overline{X\setminus A}$ which is the interior boundary in my notation. – Mittens Mar 10 '22 at 00:56
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    @SydAmerikaner: That the two axiomatizations are equivalent follows from the fact that each of them produce a topology where $\partial A=\overline{A}\cap\overline{X\setminus A}$. – Mittens Mar 10 '22 at 01:12
  • Thank you for the clarification. I understand that both yield the same topology and can follow the proofs you provided. However, I now wonder whether it can be shown that both axiomatizations are equivalent without using the topology; see my new follow-up question for details (https://math.stackexchange.com/questions/4400393/equivalent-conditions-for-set-valued-function) – Syd Amerikaner Mar 10 '22 at 18:57