Ah I'm still puzzled. So looking at the unit ball in $\Bbb{R}^3$, a $2$-sphere, any good subset will be 2 dimensional with a boundary that's a surface, @MikeMiller.
I may be really wrong, but is the reason for treating $n$-dimensional spaces as $n-1$-dimensional manifolds because topology "ignores bending", @MikeMiller? So in that respect, circles are treated like lines because you can bend a line into a circle.
Just because of the ambient space? But the dimension of your manifold shouldn't depend on the dimension of the ambient space. Dimension means "How many independent directions can I move in?"
Please don't flag things that aren't offensive. Bear in mind that not everyone has ChatJax on, and can't see what messages are beyond the LaTeX commands.
i have a question on topology: $f,g:E\rightarrow F$ two continuous function and F is hausdorff, in order to prove that this set $\{x\in E, f(x)=g(x)\}$ is closed one use that $\Delta=\{(x,x), x\in X\}$ is closed, how we get the idea to use $\Delta$ ? thank you
the idea of using $\Delta$ is considering $v(x) = (f(x), g(x))$. then the set where the two functions are equal is precisely the preimage of $\Delta$ under $v$
Some days ago, our math teacher said that he would give a good grade to the first one that will manage to draw this:
To draw this without lifting the pen and without tracing the same line more than once. It's a bit like the "nine dots" puzzle but here I didn't find any working solution.
So I h...
@Vrouvrou: Actually, rereading your question I think the approach with $\Delta$ should be used. A topological space is just a set with a topology. No notion of addition or $0$.
@Vrouvrou No, there is no notion of addition in an arbitrary topological space. Just consider the function $(f,g):E\to F\times F$. You know that $\Delta\subset F\times F$ is closed by Hausdorfness of $F$, so....
Guy's would it be ok if you could help me out understanding something? I've already asked this question but I really don't want to spam the Questions page.
No, the open sets in $E \times F$ are not quite $U_E \times U_F$ where $U_E, E_F$ are open sets in $E, F$ respectively. I mean, that's not enough for a topology. The open sets in $E \times F$ in the product topology is generated by the sets of the form $U_E \times U_F$.
Good, but again, it is not true that a neighborhood of $(x, y)$ in $E \times F$ is always going to be of the form $V_1 \times V_2$ where $V_1$ is an nbhd of $x$, $V_2$ is an nbhd of $y$. But you're close. What if $V$ is union of things of the form $V_1 \times V_2$?
Be explicit. An open set can be written as union $\bigcup U_i$ where $U_i$ are primary open sets. Thus, if $f : X \to Y$ sends each primary open set to open sets, then $f(\bigcup U_i) = \bigcup f(U_i)$ is also open as each $f(U_i)$ is open and union of open sets is open.
Hence, $f$ is open.
Thus, as you proved $P : E \times F \to E$ sends primary open sets to open sets, $P$ is open.
@Vrouvrou $f$ need not send primary open sets to primary open sets. It just sends primary open sets to open sets. Anyway, we assumed $f$ does. In our case, $P : E \times F \to E$ indeed sends primary open sets to open sets, as $P(V_1 \times V_2) = V_1$, as you noted.
We have that $$\sigma (u,v)=\gamma (u)+v\delta (u)$$ and $$K=\frac{-(\dot\delta \cdot \textbf{N})^2}{EG-F^2}$$ I want to show that if $\gamma$ is a curve on a
surface $S$ and $\delta$ is the unit normal of $S$, then $K = 0$ if and only if $\gamma$ is a line of curvature of $S$.
$$$$
Doesn't t...
Some days ago, our math teacher said that he would give a good grade to the first one that will manage to draw this:
To draw this without lifting the pen and without tracing the same line more than once. It's a bit like the "nine dots" puzzle but here I didn't find any working solution.
So I h...
We have that $$\sigma (u,v)=\gamma (u)+v\delta (u)$$ and $$K=\frac{-(\dot\delta \cdot \textbf{N})^2}{EG-F^2}$$ I want to show that if $\gamma$ is a curve on a
surface $S$ and $\delta$ is the unit normal of $S$, then $K = 0$ if and only if $\gamma$ is a line of curvature of $S$.
$$$$
Doesn't t...
