Trifis said:
@tiny-tim maybe you could reply to a semi-relevant question of mine in this thread too! I don't want to open a new one only for this question:
I am curious about how the topologists prove whether a given space possesses a specific property or not. Is there a formal, mathematical way to prove for example that M1 = {(x,y)\inℝ2 | |y2+ x2|≤ 10} is not bounded or convex but it's closed and starlike or that M2 = {(x,y)\inℝ2 | |x2-y2|>1} is not connected? Or do we just draw the sketches and decide with the eye what is what?
M
1 is actually both bounded and convex. I think you would benefit from seeing a specific example of how such properties can be proved formally, so I'll prove this claim. I should also mention that I know very little topology, so I can't answer you question about what topologists do in practice. Here we go...
Proving that M
1 is bounded is easy. Note that for (x,y)\in\mathbb{R}^{2}, \left|x^{2}+y^{2}\right| = \left\|(x,y)\right\|^{2}, so
(x,y)\in M_{1}
iff \left\|(x,y)\right\|^{2} \leq 10
iff \left\|(x,y)\right\| \leq \sqrt{10}.
Hence, by definition, M
1 is bounded. (If you're using the sup norm, use the fact that the sup norm of a point is always ≤ the euclidean norm, so the sup norms of all the points in M
1 are also ≤ √10.
To prove that M
1 is convex, let a,b\in M_{1}. We want to prove that the entire line segment \left\{a+t(b-a) : t\in[0,1]\right\} is contained in M
1. But for t\in[0,1], we have
\left\|a+t(b-a)\right\|
= \left\|(1-t)a+tb\right\|
\leq (1-t)\left\|a\right\|+t\left\|b\right\| (triangle inequality)
\leq (1-t)\sqrt{10}+t\sqrt{10} (from the fact that a and b are in M
1 and the observation made in the first proof that the points in M
1 are exactly those points with norm ≤√10.)
= [(1-t)+t]\sqrt{10}
= \sqrt{10}.
That is, \left\|a+t(b-a)\right\| \leq \sqrt{10}. Therefore, a+t(b-a)\in M_{1}. Done.