Hello, I have a little question about direct sum and direct produc of R-modules. Is the direct sum and also the direct product of a finite number of modules the same as the cartesian product (same as in equal by definition or as isomophism?) Thanks!
@ParthKohli, I just want to be sure about this: $f_n(z) = \frac {2nz -1} {z + n^2}$ is not uniformly convergent, but is only pointwise convergent to 0?
It's pointwise convergent to 0 since $lim_{n \rightarrow \infty} f_n (z) = 0$. However, I can't seem to find bounds for $f_n(z)$ within the disc, $|z| < 1$.
@DanielFischer In a general normed space, how would you go about proving that any convex function is continuous ? It seems the real argument of differentiability doesn't adapt
@LeGrandDODOM I guess that would imply continuity. But I don't know. $$\psi_x \colon h \mapsto \lim_{t \downarrow 0} \frac{f(x + th) - f(x)}{t}$$ should define a sublinear functional. If $f$ is bounded on a neighbourhood of $x$, $\psi_x$ should be bounded on a neighbourhood of $0$, so continuous. Thus the subdifferential isn't empty, and I think that should imply continuity of $f$ at $x$. Dunno if all that works, though.
Oh, good — but then you should have caught my sarcasm.
Here's a good puzzle for you guys (you may already have thought about this). It actually came up in a question on main, but it also came up in the integration portion of my course about 4 years ago. ... Give me a bounded open set in the plane whose boundary (in the point-set sense) has positive measure.
(My glib response to my students who asked on the video was to mumble space-filling curve, but I told them to come talk to me in office hours about a more direct construction.)
I went back and checked when the question came up on main. It occurred when I was discussing the theorem that a function with discontinuities a set of volume 0 is (Riemann-) integrable.
Your answer is better than the construction Alessandro will do (:)) because the closure of your open set is an interesting compact set.
Probably, @Alessandro, although I don't think you need a product of two Cantor sets. There's an easier construction, yet, although it has, I suppose, a thick Cantor set flavor.
The annoying thing with this example (and hi, @DanielF) is that the closure is the whole square. So an example where the closure is a proper subset would be more interesting. But, anyhow ...
Here's a theorem Bob talked about yesterday. Suppose you have a knot in $S^2 \times S^1$ that intersects one of the 2-spheres transversely in a single point. Prove that it's the unknot.
if the knot intersects a $2$-sphere $S^2\times\{y\}$ exactly once, then there should be a nbhd $Y$ of $y$ for which it intersects each $2$-sphere of $S^2\times Y$ exactly once, so we should be able to "disconnect" the knot from there and look at a path from one boundary component to another of $S^2\times (S^1\setminus Y)$. then we can contract each boundary to a point and we're looking at a loop in $S^3$
Thicken then S^2 up into an S^2 x [0, 1] so the knot comes at S^2 x 0 and goes out from S^2 x 0. Thicken up it into a tubular neighborhood and look at boundary. That's an embedded S^2 (two S^2's tubed togather) in the S^2 x S^1, right? And it's complement is S^2 x S^1 minus a ball...
Think of, like, a proper trefoil (trefoil knot with two ends extending off to infinity in R2 x [-1, 1]) and then think of identifying R2 x {-1} with R^2 x 1 by identity and one point compactify fiberwise to get it inside (R2)^* x S^1 = S^2 x S^1. Why's this not a nontrivial knot (it surely hits an R^2 x {p} transversely at one point)?
Oh so I guess I understand what happens after fiberwise compactification. You get to move a "handle" of a crossing up to the R^2 it hits transversely, and then make it lie flat, and then move it so it goes round the point at infinity and comes back. I bet moves like that would resolve the knot
Sounds hard to actually prove to be working
Sounds like it should flip an overcrossing-undercrossing pair into an undercrossing-overcrossing pair
Wouldn't finitely many moves like that unknot stuff?
I meant one which hits an S^2 transversely at a point (because you can move the "handle" overcrossing the stuff which goes inside S^2 and then slide it around the sphere to make it lie behind the stuff coming out of S^2 from the other side - but those are the same things so you flipped overcrossing/undercrossing)
There's a much more straightforward phrasing of your proof. Straighten the knot to be constant (on S^2) near t. Cutting that open, we get a "long knot" in S^2 x R = (R^3 \ 0) that statts from 0 along the z-axis and ends at infinity going up the z-axis.
Then this is like a lightbulb dangling from a string, and you can unknot that without moving the lightbulb or the ceiling - just pull knots down under the lightbulb and you can change overcrossings to undercrossings.
This was your operation of pulling things across infinity.
(the negative z-axis corresponds to infty times R)
Let $\tau$ be a familliy from $\mathcal{P}(\mathbb{R})$ consisting by $\mathbb{R}$ and any set such that it's complement containes $[x_0]$ where $x_0>10$
!. give the nature of $]9,10]$ and $[9,10[$
----> I think that these two sets are open
2. Distinguish the open sets
---->I don't understand this question
3. give a basis of $\tau_{\mathbb{N}}$ i don't know how to do ?
I think that exercise is very poorly phrased, the whole of $\tau_{\Bbb N}$ is obviously a basis for itself, I don't know if they expect a smaller one as answer
@Vrouvrou can't really tell you much without solving the exercise in your stead... Are there sets containing $[x_0]$ with a finite complement in $\Bbb R$?
