# Motivation behind point set topology [closed]

Why should I study point-set topology?

What initially interested me in topology was the pop-sci rubber sheet stuff or coffee cup-donut stuff or proving fundamental theorem of algebra using curves but now here I'm stuck between distinguishing between $$T2.5$$ and $$T2.500001$$ spaces with loads of weird counterexamples and all that.

Also, I don't find any applications of point-set topology mentioned in Munkre's book either except for maybe finding weird counterexamples in real analysis but then honestly I find that to be very boring too (maybe something is wrong with my mathematical interests).

## closed as primarily opinion-based by uniquesolution, Lord Shark the Unknown, Eevee Trainer, Javi, José Carlos SantosApr 7 at 11:04

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• I have the exact opposite attitude, I find point-set topology interesting but I find algebraic topology with its fundamental groups and surface classification theorems quite boring. Different strokes for different folks. – Keshav Srinivasan Apr 3 at 5:55
• @KeshavSrinivasan What are the topics in point set topology that you find interesting ? Then maybe I can use that to motivate myself. – cdt Apr 3 at 6:01
• Well, what I find interesting are how other kinds of structures, like metrics, uniformities, and bornologies, interact with the topological structure of a space. You can check out my questions here: math.stackexchange.com/… – Keshav Srinivasan Apr 3 at 6:29
• To do what you describe in the first paragraph you don't have to study T2.5 spaces. You just need to be aware that this theory exists somewhere in the background and forever work with simplicial complexes and/or manifolds. – Michal Adamaszek Apr 3 at 8:42
• You are not obliged to learn "mathematics and stuff". – Yves Daoust Apr 3 at 9:15

One of the points of studying point-set topology is that it prepares you for algebraic topology (what you seem to refer to when you say for instance "donut coffee cup stuff")

(Of course point-set topology is interesting in its own right, but clearly alxchen isn't interested in it, so this is an answer based off this assumption)

A bunch of algebraic topology can be done without even thinking of point set topology (you can work in simplicial sets, or model categories etc.) but at least at the foundations of it you will find yourself needing some point set topology and some understanding of pathologies. For instance, say you're interested in manifolds, and you have a manifold $$M$$ which you want to quotient by the action of a group $$G$$. Then point-set topology will enter the door by telling you what sort of action you need in order for the quotient to be reasonably well-behaved : knowing the pathologies of point-set topology can help you avoid them.

Similarly, say you're not interested in the difference between $$T_1$$ and $$T_2$$. Then the real line with two origins sounds perfectly fine for you; but it happens to be $$T_1$$ and not $$T_2$$, so a lot of manifold theory that has been developped specifically for $$T_2$$ manifolds fails there.

Another example is with CW complexes, one of the most important types of spaces for algebraic topology. They seem pretty nice and pretty far from all the point-set pathologies, but when they're infinite dimensional, pretty wild things can start happening, and you need to know a few things about point-set topology to avoid problems.

You also need to know the basics of point-set topology at the beginning to prove that certain homotopies or paths are continuous, to find the right assumptions that make covering space theory work, etc.

My last point will be that point-set considerations are important to know with what kind of space you want to work with to do the more advanced algebraic topology stuff : for instance the exponential law $$\mathbf{Map}(X\times Y, Z)\cong \mathbf{Map}(X, Z^Y)$$ only works under specific hypotheses on the spaces involved, and since it's a pretty nice property, if you want it, you'll want to find the appropriate kinds of spaces that work the best in your context.

(Personnally I find point-set topology super fun -except when it gets in the way of my algebraic topology- so I could have said a bunch of stuff about that too; but different people have different tastes and my point here was not to convince you that point-set topology is cool, because you might just not be into it)

Those weird counterexamples are, from a certain standpoint, the reason to love point-set topology: they provide you with a much better understanding of how things you think you understand can go wrong. What point-set topology really does is explain structure and lets you evaluate new objects (and sometimes old ones) to see where their structure is familiar and where it's odd, and that can help lead you in new directions.

But let's take an example: the Baire Category Theorem is an abstract result that says that if sets are nowhere dense (i.e. the interior of their closure is empty) then a countable union of them has the same property. A Baire-zero function is just a continuous function, and a Baire-one function is the pointwise limit of continuous functions.

To be less abstract: let $$f_n : [0,1] \rightarrow [0,1]$$ be defined by $$f_n(x)=\left\{ \begin{eqnarray} 0 & \mbox{if} & x \in \{0,1\} \\ 1 & \mbox{ if } & \frac{1}{n} < x <1-\frac{1}{n} \\ nx & \mbox{ if } & 0 < x < \frac{1}{n} \\ n(1-x) & \mbox{ if } & 1-\frac{1}{n} so the $$f_n$$ are trapezoids. Their limiting function is Baire-one by definition (since they're all continuous) and is $$f:=\chi_{[0,1]}-\chi_{\{0,1\}}$$ where $$\chi_A$$ is the characteristic function of a set $$A$$ -- and is clearly discontinuous. (We can, as you've probably guessed, extend this to find Baire-two functions that are the pointwise limits of Baire-one functions, etc. And... there are functions that do not belong to any finite Baire-class!)

Those kind of functions will be familiar to you already as a source of counterexamples (for things like uniform convergence of Lebesgue integrals, for example). Now, we can use the Baire category theorem to show that Baire-one functions have a point of continuity in every closed interval that it's defined on, which tells us something interesting: the limit of pointwise continuous functions can't be too discontinuous. (That said, those discontinuities can form quite a large set in measure-theoretical terms; see https://math.stackexchange.com/a/112133/13130 ).

We know another kind of thing that can't have too many discontinuities: derivatives of differentiable functions. And indeed, we can show (relatively easily) that derivatives must be Baire-one functions. Now, I wrote $$f$$ deliberately as a difference of characteristic functions, because, going all the way back to point-sets, the characteristic functions of $$F_\sigma$$ sets (countable unions of closed sets) are exactly the Baire-one functions. Better still, the discontinuity set of a derivative must be a first-category $$F_\sigma$$-set. So we get a (probably surprising) result: you can know a lot about derivatives by looking at the countable unions of closed sets!

[Thanks to Dave L. Renfro for the link above, and for taking the time to correct my careless original post!]

• Some of this needs editing/fixing. For example, the part about "if sets are scattered enough (so much so that they contain no open interval about any of their points)" seems misstated or misleading, and Baire one functions (even derivatives) CAN have uncountably many discontinuities (in fact, the set $D$ of discontinuities of a Baire one function can have the property that for each interval $I$ the Lebesgue measure of $D \cap I$ is equal to the length of $I).$ – Dave L. Renfro Apr 3 at 10:47
• @DaveL.Renfro Hi, and thanks. You're right, the Baire-one functions can't have a dense set of discontinuities. Can I ask what the problem is with me saying that a set of first category contains no open interval though? – postmortes Apr 3 at 10:52
• Baire-one functions CAN have a dense set of discontinuities. Regarding "if sets are scattered enough (so much so that they contain no open interval about any of their points) then a countable union of them has the same property, note that each of $\mathbb Q$ and ${\mathbb R} - {\mathbb Q}$ has this property but their union doesn't have this property. – Dave L. Renfro Apr 3 at 11:15
• For possibilities of what the discontinuities of a Baire one function can be, see this answer along with this other answer. – Dave L. Renfro Apr 3 at 11:29
• @DaveL.Renfro thank-you again for your help and clarifications. I'm actually studying Bruckner at the moment -- clearly I need to work harder at it! – postmortes Apr 3 at 12:00