Recent content by sapporozoe

Thanks. This case is a bit tricky. We are asked to find the fixed points under this correspondence...As you know, the original correspondence is not convex-valued and has no fixed points. Then the convex hull is convex-valued and by Kakutani's theorem there exists at least one fixed point. I...

First, I wonder whether I can put the post here... Given X=[0,1]^2 a(x)={y in X:||y-x||>=1/4} b(x)is the convex hull of a(x). Identify the set of fixed points. My answer is 3/4>=x>=1/4, 3/4>=y>=1/4, but I am not sure... What if we have a(x)={y in X:||y-x||>=1/2}? (My answer is...

X=[0,1]^2 a(x)={y in X:||y-x||>=1/4} b(x)is the convex hull of a(x). Identify the set of fixed points. My answer is 3/4>=x>=1/4, 3/4>=y>=1/4, but I am not sure... Thanks.

i almost get it:) Thanks.

when i wrote down the previous post, i was worried whether i made my question clear:) The statement is true because we exclude cases like: f(x)=tanx, x in (-pi/2, pi/2). We can exclude such cases as we require the domain of f(x) is Rn. So my question becomes: if we say "f is from Rn to...

Thanks for your patience:). But the following example tells my confusion... Perhaps we are familiar with a statement like this:＂if f:Rn to Rm is continuous and B in Rn is bounded, then f(B) is Rm is bounded". We know that boundedness is not preserved under continuous mapping. But this...

thank you. but do we usually take Rn as the domain and Rm as the range when we say "f is from Rn to Rm"? I am a bit confused about this...

great! thanks...

but... but the range of this f is not R, is it?

If f from R to R is continuous, does it then follow that the pre-image of the closed unit interval [0,1] is compact? -At first I thought of a counterexample like f=sinx but it seems that its range is not R. So will the answer be yes? And how can we prove it? Will the preimage have to be...