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I've noticed one classical way of defining certain topologies is to define them as the "weakest" (or coarsest) topology such that a certain set of functions is continuous. For example,

The product topology on $X=\prod X_i$ is the weakest topology such that the canonical projections $p_i : X\to X_i$ are continuous.

The weak*-topology on a Banach space $X$ is the weakest topology such that the evaluation map (the natural isomorphism from $X$ to $X^{**}$, $J(x)(\phi) = \phi(x)$) is continuous.

I have a very poor intuition behind what these "mean". When I look at the definition of a topology, the ones that make the most sense to me are the ones in which the open sets (or at least a base of open sets) are explicitly constructed, as in the Euclidean topology (or a general metric space topology, or a norm induced topology).

I get a little stuck when the definition of the topology is given in some "abstract" sense, where the open sets are "chosen" to satisfy a certain other property. How am I supposed to visualize the open sets in these spaces, or work with them?

If $\tau$ is the weakest topology on $X$ such that $f : X\to Y$ is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets in $Y$ under $f$? This follows directly from the definition of a continuous function on topological spaces. Is this always the coarsest topology?

Furthermore, what benefit do these topologies provide? What interesting, and potentially theoretically useful, properties do they possess? Why should I care about them?

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    $\begingroup$ Here weakest means the topology with the fewest open sets. So you are right, you have to throw in preimages of the open sets under the given functions and everything that the axioms of a topology then force you to. $\endgroup$ – Cheerful Parsnip May 8 '18 at 22:33
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If $τ$ is the weakest topology on $X$ such that $f:X→Y$ is continuous, is it correct to imagine a base for the open sets to be the preimage of all open sets under $Y$? This follows directly from the definition of a "continuous" function. Is this always the coarsest topology?

Yes. For $f$ to be continuous, you need the topology on $X$ to contain all preimages of open sets through $f$. The topology induced by a family $\mathcal T$ of functions is generated by $$ \{f^{-1}(E):\ f\in\mathcal T,\ E\subset Y\ \text{ open }\}. $$ Some reasons why one cares about these topologies are

  • They often appear naturally, as when considering duals and preduals of normed spaces;

  • In several cases the topology is coarse enough that some interesting sets become compact (for instance the unit ball in a Banach space, see the Banach-Alaoglu Theorem).

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    $\begingroup$ They are the smallest $\sigma$-algebra that contains the open sets. What I'm saying is that the $\sigma$-algebra generated by the preimages of open sets, is the set of preimages of Borel sets. $\endgroup$ – Martin Argerami May 8 '18 at 22:36
  • $\begingroup$ Thanks for the response! I'm quite glad these "coarsest" topologies actually have a pretty simple explicit definition as they show up so often (especially the product topology). Unrelated: I must say I've never heard of preduals before today and from the name alone they sound absolutely crazy. $\endgroup$ – user3002473 May 8 '18 at 22:42
  • $\begingroup$ Not really, when you get familiar with them. When you have a dual of something, the something is a predual... When $V$ is a vector space and $V^*$ is it's dual, you have a very natural topology, which is that of pointwise convergence; it is precisely the coarsest topology that makes each $v$ continuous when seen as a map $f\longmapsto f(v)$. We call this the weak$^*$-topology. $\endgroup$ – Martin Argerami May 8 '18 at 22:46
  • $\begingroup$ One more thing Martin: I think the $\mathcal T_{\mathcal F}$ you give is just the subbase of the weak topology. $\endgroup$ – Aweygan May 8 '18 at 22:48
  • $\begingroup$ Let's see if I say my $n^{\rm th}$ stupid thing today: if I pull back a topology through preimages, I get a topology since preimages preserve arbitrary unions and intersections. $\endgroup$ – Martin Argerami May 8 '18 at 23:04
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"weakest"

Fix a space X, and let Top(X) be the set of topologies on X. That is, an element T of Top(X) can be thought of as a subset of the power set P(X) (satisfying axioms). Now, Top(X) naturally forms a poset under $\subseteq$: that is, if S and T are elements of Top(X), you can think of S as "smaller than" T if $S\subseteq T$, i.e. all S-open sets are also T-open (but not necessarily conversely). The indiscrete topology is the "smallest" element of this poset, and the discrete topology is the "largest" element.

The words "coarse" and "fine" are synonyms in this context for "small" and "large" respectively. That is, the indiscrete topology is the coarsest possible topology (it has very few open sets; it smooshes all points of X together into one big open set, so that you can't tell the points or subsets apart with open sets), and the discrete topology is the finest possible topology (it has lots of open sets; you can tell lots of things apart). If X is e.g. $\mathbb{C}^n$, then you could also define the Euclidean topology, or the (weaker) Zariski topology.

Note the following: if S and T are both elements of Top(X), then $S\cap T$ is too. (Prove it!)

The "weakest" topology subject to some conditions, then, is the smallest possible topology that satisfies those conditions. Put another way, it's the intersection of all the topologies that satisfy those conditions. Put yet another way, if you are happier with algebra than topology, you might like to think of this as the topology generated by some open sets.

$f: X\to Y$ is continuous if, for every open U in Y, the set $f^{-1}(U)$ is open in X. So the weakest topology on $X$ such that $f:X\to Y$ is continuous is the smallest one that contains $f^{-1}(U)$ for every open U in Y. That's the intersection of all those T in Top(X) such that $f^{-1}(U)\in T$ for all open U in Y. And so on.

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  • $\begingroup$ I actually like this definition of the weakest topology on $X$ such that $f$ is continuous, as it reminds me of the construction of the "interior" of a set as the union of all its open subsets. $\endgroup$ – user3002473 May 8 '18 at 22:39
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    $\begingroup$ @user3002473 ... or the span of some vectors as intersection of all containing sub-sapces, or the subgroup generated by some set as intersection of containing subgroups, or the naturals as smallest inductive set, or ... it's a recurring theme $\endgroup$ – Hagen von Eitzen May 9 '18 at 4:57

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