Mathematics - 2014-11-27 (page 3 of 3)

11:31 PM

@TomCruise You want the math? You can't handle the math.

Tom Cruise

Did you order the code red!!!!!!

I want to show that the set of functions bounded by M is nowhere-dense in $C[0,1]$ with the metric $d(f,g)=\int_0^1|f(t)-g(t)|dt$

@TomCruise You want solutions?

@Studentmath Yes.

@PedroTamaroff Hi!!! I want to show that if $V$ is an algebraic set of $K^n$, $V$ is irreducible iff I(V) is a prime ideal of $K[x_1, x_2, \dots, x_n]$

So far, I showed that if $V$ is not irreducible, $I(V)$ is not a prime ideal.. How could I continue?

You can't handle the solutions.

11:37 PM

Now, I think it's enough to prove that every open ball in $C[0,1]$ has some ball disjoint with that set of functions - right?

@evinda Where are you stuck?

@PedroTamaroff I have shown that if $V$ is not irreducible, then $I(V)$ is not a prime.
Do we conclude from that, that if $I(V)$ is a prime ideal, then $V$ is irreducible?

@Pedro is it a smart idea to find the closure of the set of functions bounded by M for starts?

@evinda What do you think?

@Studentmath Let me read your problem.

@evinda Did you know that $V(\mathfrak a\mathfrak b)=V(\mathfrak a)\cup V(\mathfrak b)$?

@Pedro cheers, I think I will find the closure, and then prove $Int(ClA)=\emptyset$

11:47 PM

@Studentmath Are yo leaving?

You always leave before I can solve your problems.

Haha, I actually thought to walk the dog, but I will delay it :P

@PedroTamaroff I think that we have shown that if $I(V)$ is a prime ideal, then $V$ is irreducible, am I right?

I tried to prove this, but then I got stuck..That's what I have tried:

$x \in V(IJ) \leftrightarrow (f_i \cdot g_j)(x)=0$, where $f_i \in I$ and $g_j \in J$

$\leftrightarrow f_i(x) \cdot g_j(x)=0 \leftrightarrow f_i(x)=0 \text{ OR } g_j(x)=0 \leftrightarrow x \in V(I) \text{ OR } x \in V(J)$ $\leftrightarrow x \in V(I)\cup V(J)$

So, $V(IJ)= V(I)\cup V(J)$.

I think it has to do with your wake-and-able-to-help hours meeting my sleep hours

@Studentmath So, let's just rewrite the definitions.

To prove a set $A\subset X$ is nowhere dense, you ought to show that its closure has empty interior,

Yep

11:50 PM

Can you write this in terms of balls?

@PedroTamaroff Also, how could we prove that if $V$ is irreducible, then $I(V)$ is a prime ideal?

@Pedro well, I think $A\subset X$ is nowhere dense iff every open ball in $X$ has as a subset a ball that is disjoint with $A$

@Studentmath Yes.

So, in this situation, given a ball $B(f,\varepsilon)$, you're supposed to find $B(g,\delta)$ inside this ball such that each $g$ is not bounded above by $M$. I guess it might be useful to think about aiming at a contradiction, perhaps.

Ahah, that should make it much easier - one question though, that I struggle with - how do I show that $B(g,\lambda ) \subset B(f,\epsilon )$?

It's obviously true if $g=f$ and $\lambda < \epsilon$, but if $g\neq f$? I just have to show that every function at that distance from $g$ is at $\epsilon$ distance from $f$ by definition, right?

@Studentmath Are you considering the set of functions strictly dominated by $M$?

Or just at most $M$?

11:57 PM

At most, $\le$

Isn't that closed? Then it suffices you show $A$ has empty interior.