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I am not sure if this question actually makes sense, and I am also not sure if the details of this question are entirely correct, so any corrections would be greatly appreciated!

In my math class, we learned that every inner product space is a normed space and that every normed space is a metric space. Is it true that every metric space is a topological space? But, the converses of these may not be true. Maybe I should replace "is" with "may induce a"?

But since these are getting more and more general, is there something that a topological space may induce or something that is more general than a topological space?

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  • $\begingroup$ Well, there is "set," of course. There's also the notion of a "locale" or a "point-less topological space." en.wikipedia.org/wiki/Pointless_topology $\endgroup$ – Thomas Andrews Apr 2 '17 at 5:01
  • $\begingroup$ Topological spaces can be defined by the Kuratowski closure axioms: (i) $X\subseteq\overline X;$ (ii) $\overline\emptyset=\emptyset;$ (iii) $\overline{X\cup Y}=\overline X\cup\overline Y;$ (iv) $\overline{\overline X}=\overline X.$ The more general closure spaces are defined by the axioms (i)–iii), dropping the idempotent law $\overline{\overline X}=\overline X$ from the axiom set. On another note, I think "syntopogeneous" spaces are supposed to generalize topological spaces, but I have no idea what they are. $\endgroup$ – bof Apr 2 '17 at 5:09
  • $\begingroup$ While there are spaces more general than topological spaces, I think that most people doing mathematics would say that a topology is one of the foundational structures you can put on a set. Bourbaki list three: 1) order, 2) topology, 3) algebraic operations. As such, a topological space is itself not much more than a blank slate, as there are different ways to mix and match properties on which a topological theory may be built. In this sense of mixing and matching properties, this is more or less where the buck stops. $\endgroup$ – Alfred Yerger Apr 2 '17 at 5:11
  • $\begingroup$ Every topological space $X$ can viewed of as the category of open sets in the topology on $X$. You can retain some geometric feel to this if you also consider this category as a site, where the Grothendieck topology is defined by taking the coverings of an object $U\subseteq X$ to be collections of the form $\{U_i\to U\}$ with $\bigcup_i U_i = U$. $\endgroup$ – Stahl Apr 2 '17 at 5:33
  • $\begingroup$ (ctd) Not every category comes from a topological space in this way, and not every site is induced by an honest topological space in this way either. Categories are of course extremely general; sites have some geometric/topological flavor to them. $\endgroup$ – Stahl Apr 2 '17 at 5:34
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I will try to answer your question in elementary terms. You have seen that every normed space is a metric space (but the converse is no true) and that every metric space is a topological space (but the converse is no true). Now you are asking if there is some kind of space, let's call it a ??-space such that every topological space is a ??-space (but the converse is no true).

Of course, you could just take a ??-space to be a set, but there is a slightly more subtle answer. You could take as ??-space a preordered set. Let me first recall that a preorder is a reflexive and transitive relation. It turns out that every topological space is naturally equipped with a preorder, called the specialization preorder, which can be defined as follows: $$ x \leqslant y \iff \overline{\{x\}} \subseteq \overline{\{y\}}, $$ where $\overline{S}$ denotes the closure of a set $S$. However, you might be disappointed by this answer, because for an Hausdorff space (also called $T_2$-space), the specialization preorder is the equality relation. On the positive side, it the space is Kolmogorov (also called $T_0$), then the specialization preorder is a (partial) order.

In the language of categories, I just described the forgetful functor from the category of topological spaces to the category of preordered sets.

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Pretopology spaces, nhood spaces, precloser spaces. A precloser pcl, operator lacks plc plc A = pcl A.

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  • $\begingroup$ Your answer is not very good as it is. You should expand it a little bit, explain the concepts with more details, maybe give examples or references. See also math.stackexchange.com/help/how-to-answer $\endgroup$ – Arnaud D. Apr 2 '17 at 11:07

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