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)[itex]\in[/itex]ℝ2 | |y2+ x2|≤ 10} is not bounded or convex but it's closed and starlike or that M2 = {(x,y)[itex]\in[/itex]ℝ2 | |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 [itex](x,y)\in\mathbb{R}^{2}[/itex], [itex]\left|x^{2}+y^{2}\right| = \left\|(x,y)\right\|^{2}[/itex], so
[itex](x,y)\in M_{1}[/itex]
iff [itex]\left\|(x,y)\right\|^{2} \leq 10[/itex]
iff [itex]\left\|(x,y)\right\| \leq \sqrt{10}[/itex].
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 [itex]a,b\in M_{1}[/itex]. We want to prove that the entire line segment [itex]\left\{a+t(b-a) : t\in[0,1]\right\}[/itex] is contained in M
1. But for [itex]t\in[0,1][/itex], we have
[itex]\left\|a+t(b-a)\right\|[/itex]
[itex]= \left\|(1-t)a+tb\right\|[/itex]
[itex]\leq (1-t)\left\|a\right\|+t\left\|b\right\|[/itex] (triangle inequality)
[itex]\leq (1-t)\sqrt{10}+t\sqrt{10}[/itex] (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.)
[itex]= [(1-t)+t]\sqrt{10}[/itex]
[itex]= \sqrt{10}.[/itex]
That is, [itex]\left\|a+t(b-a)\right\| \leq \sqrt{10}[/itex]. Therefore, [itex]a+t(b-a)\in M_{1}.[/itex] Done.