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When defining a topology via the open set definition, why do we not insist that $\varnothing \neq X$?


The open set definition of a topology includes the following things:

  1. $\varnothing$ is an element of $\tau$.
  2. $X$ is an element of $\tau$.
  3. If $\mathcal{F}$ is a family of elements of $\tau$, then $\cup \mathcal{F}$ is an element of $\tau$ too.
  4. If $A$ and $B$ are elements of $\tau$, then $A \cap B$ is an element of $\tau$.

Topologies cannot be turned into a first-order theory, but conditions (1) and (2) are reminiscent of the existence of $0$ and $1$ in fields, and in that setting we insist that $0 \ne 1$ which rules out interesting psuedofields like the field with one element.

The above analogy is extremely loose, but I'm curious why we don't do the analogous thing for topologies and insist that $\varnothing \neq X$.

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  • $\begingroup$ When "topology" was originally being defined (by many different ways, not always equivalent) people didn't worry about empty set issues such as this. And later when this might have been considered (by textbook authors in the 1930s-1950s), I suspect the default was to not assume "nonempty" since it wasn't strictly needed. It is EXTREMELY UNLIKELY that these authors gave any consideration to issues with first-order theories and $0 \neq 1$ in fields (and why consider only fields ... $0=1$ is possible in rings). $\endgroup$ Commented Jun 8, 2021 at 16:20
  • $\begingroup$ This comment might also be of interest. $\endgroup$ Commented Jun 8, 2021 at 16:26
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    $\begingroup$ I would counter: for what topological reason should it be excluded? $\endgroup$
    – Randall
    Commented Jun 8, 2021 at 16:39
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    $\begingroup$ Also, the homology of pairs becomes extremely annoying and clumsy to state if the empty set is somehow forbidden to carry a topology. $\endgroup$
    – Randall
    Commented Jun 8, 2021 at 16:41
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    $\begingroup$ In many geometric situations, you want to look at subspaces of an ambient space and look at their intersection. Sometimes, subsets are disjoint, i.e. the intersection is empty. Yet, you still want to talk about that intersection as a geometric object. The empty topological space does this. $\endgroup$
    – Thorgott
    Commented Jun 8, 2021 at 17:35

2 Answers 2

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Maybe because the empty set (considered as a topological space) is the initial object of the category of topological spaces.

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I guess that you might be correct, that if empty set is allowed as a topological space, then it makes sense to also allow the zero ring as a legitimate commutative unital ring. But, a field should still be defined as having $0 \neq 1$. So the zero ring should not be considered a field. This is analogous to how $1$ is not a prime number. For example, you want unique decomposition into primes, and if you set $1$ to be a prime number you don't have it (but if not, then very elegantly $1$ decomposes to the empty product of primes, so everything is well). So fields correspond to "possible atoms out of which a spectrum is built", and they should be non-trivial.

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    $\begingroup$ There are similar things to this "field must not be trivial": an irreducible representation is a non-zero representation such that... A connected topological space is a non-empty topological space such that... The last example relates to your original question. The empty space should be considered not connected. $\endgroup$
    – Sasha
    Commented Jun 8, 2021 at 16:25
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    $\begingroup$ In every reference I’ve ever seen the empty space is connected. There is no such proviso. $\endgroup$
    – Randall
    Commented Jun 8, 2021 at 16:43
  • $\begingroup$ Also, this seems to repeat OP’s question as opposed to providing an answer to it. $\endgroup$
    – Randall
    Commented Jun 8, 2021 at 16:49
  • $\begingroup$ @Randall You can also see discussion here: ncatlab.org/nlab/show/empty+space Anyway, I think it is correct to think of empty space as not connected. Connected space is a space whose set of connected components is "contractible". Contractible is something that is equivalent to a point, and thus non-empty. I guess this is "correct", versus "common". $\endgroup$
    – Sasha
    Commented Jun 8, 2021 at 17:23
  • $\begingroup$ To repeat, I think that the idea is that "empty" structures are generally welcomed and useful to consider in the arsenal of structures. However, "irreducible constituents" should not be "empty" (and "empty" should be thought as the empty "integration" of "irreducible constituents"), and so commutative unital rings can be trivial, but fields should not be allowed to be trivial (because fields are what describes "points", i.e. "irreducible constituents"). $\endgroup$
    – Sasha
    Commented Jun 8, 2021 at 17:28

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