Every real polynomial of two variables defined on the real plane R^2 and bounded from below attains its minimum

Every real polynomial of two variables defined on the real plane R^2 and bounded from below attains its minimum.

is this supposed to be remarkable?

It's remarkably false.

Proofs?

(1-xy)^2 + x^2

Bounded from below just means that the surface never goes below that point.

The surface could still approach that point arbitrarily closely and thus never have a minimum.

O, yea. Is there some variation on OPs claim? Or he just said stupid shit?

Every subspace of a completely metrizable space is completely metrizable.

Every such function on a bounded subset of R2 will achieve its minimum.

>Is there some variation on OPs claim?

Do you mean for it to be right? Well, yes. It is true for real polynomials of one variable.

Babby proof:

The only polynomials to be bounded from below are even degree polynomials.

Every (and only) even degree polynomials are either bounded from below or above.

Even degree polynomials have either minima or maxima, never both.

Consider some arbitrary even degree polynomial that is bounded from below.

Take its derivative, which yields an odd degree polynomial.

Every odd degree polynomial has at least one real root.

These roots are either maxima or minima. But they cannot be maximum because if an even polynomial has a maximum then all of its critical points would be maxima and if that were the case then one of them would have to be a global maximum but then that global maximum would imply that the function is bounded from above.

But we already supposed the function was bounded from below, and even polynomials cannot be bounded from both sides.

So all of those critical points must be minima and therefore one of them must be global, so such polynomials always attain their minimum.

no, every closed subset :)

Has a minimum at (0, 0).

Sure thing bro

>But they cannot be maximum because if an even polynomial has a maximum then all of its critical points would be maxima
that's not true, and even degree polynomials can have local maxima (ex: (x-1)(x-2)(x-3)(x-4))
What you need to say is precisely that if a nonconstant polynomial has at least two critical points, then they can't all be maxima or minima (which is pretty clear if you draw the situation)

take x close to 0 and y=1/x faggot, the function approaches 0 but never reaches this value

x-y

Is it a polynomial? Dumbass

The set A = [0, infinity) x [0, infinity) is closed yet the function f with f(x,y) = -x for all (x,y) in A has no minimum.

Let X denote a compact and convex subset of euclidean space R^n. This implies that the set of its extreme points is closed.

You can't read

>stealing fake theorems from a fake math page on Facebook

heck off dude

I messed up my math. You are quite right; (1-xy)^2 + x^2 has an infimum of zero, but no minimum.