I'd parse it a bit differently, though. $X \cup Y$ in itself is not determined by $X$ and $Y$. You have an ambient space $W$ which is $X \cup Y$ where $X, Y$ are both closed or both open in $W$.
There's no way to make arbitrary union out of two random topological spaces
But I am sure you're aware of this. Just a heads-up.
@0celo7 Balarka meant "union" in the sense that they have non-empty intersection (because you were talking about $X\cap Y$). Of course the disjoint union is defined.
You have to identify $X$ and $Y$ along something. I can make a million different topological spaces out of a cantor set and a 8-dimensional torus. More information is needed
Take the thing I wrote: $[0, 2] = (0, 1) \cup [1, 2) \to \{0, 1\}$ which maps $[0, 1]$ to $0$ and $[1, 2)$ to $1$. Take preimage of $\{1\}$, which is open. You get all of $(1, 2]$, which is not open in $(0, 2)$.
It is upon you to figure out what you messed up. I'm off to sleep.
Good catch there. I was confused about the counterexample I wrong down, so didn't pay attention to your thing well enough. Your set theory is correct, sorry about that. Just not the topology.
@ACuriousMind That may be, but I have to look up the definitions every time I solve a problem in topology! I totally feel Hitler https://www.youtube.com/watch?v=SyD4p8_y8Kw
You're just one and a half year late posting that ;)
In other news, there are 34 hits for "Hitler" in the chat log of this chat
I wonder whether that is above or below the SE standard
@user50213 I don't know how you got a difficult PDE out of that. Choose radial coordinates and go to the frame where one of the bodies is at rest. Then you have (ignoring constants) $\frac{1}{r^2} = \ddot{r}$, which is not a hard equation to solve.
Wait
It can be a hard equation to solve, it's called the Kepler problem
@user50213 There's an easier way to get the time to collision, though: By energy conservation, you can express momentum as a function of radius. So you can express velocity as a function of radius, and then you use $T = \int\mathrm{d}t = \int\frac{\mathrm{d}t}{\mathrm{d}r}\mathrm{d}r$ where this integration should be easier than the differential equation (the expression for velocity/its inverse $\frac{\mathrm{d}t}{\mathrm{d}r}$ is not that ugly)
The only topologist who usually gives that lecture did the intro analysis, which is three consecutive lecture, so he couldn't do it for one and a half year and now he's starting with algebraic topology again which I already took
Suppose $\{c_i\}$ had the accumulation point $c$. Then $M^c$ is compact by hypothesis. Since all critical points are nondegenerate, we can form an open cover of $M^c$ by taking neighborhoods of the separated critical points $c_i<c$, of which there are infinitely many. This open cover has no finite subcover, which is a contradiction.
I don't think that's right
No, I don't buy that. What if it accumulates from above?
That means that if c is an accumulation point, then the infinitely many c_i must lie below it, because as soon as one $c_n$ lies above it, you have that no other $c_i$ after that can lie in $[c-\epsilon,c+\epsilon]$ for $\epsilon < c_n - c$.
So the only way every neighbourhood of c can contain infinitely many points is by them lying below it
@0celo7 I'm not sure if this is shown at this point, but $M^a$ would be a compact manifold that has a function without critical points on it, which can't exist.
"direct limit" is an algebraic/categorial construction
I think no one understands those notions of limit the first time they read the definition, they take a bit getting used to (at least I took my time getting used to them, and most people I know)