Oh my it's getting really funny! Yeah, but the fact is that both sets are in the ball with radius 51, and so their union is bounded. I overlooked the value of the constant K, which should of course be the diameter of the (larger) ball, i.e. 102.
I "repaired" the proof. It should be OK now...
I should have made it more explicit. Here is the complete version:
Let A and B be bounded. Take any point z_0\in X, such that for all x\in A we have d(x,z_0)\ge\sup_{u\in A}d(x,u) and for all y\in B respectively d(y,z_0)\ge\sup_{v\in B}d(y,v). Then with K_A:=\sup_{x\in A}d(x,z_0) and...
Well, I realised this at the very beginning, but it's now that I can write it down:
Let A and B be bounded. W.l.o.g assume that \delta (A)\le\delta (B). By assumption there are numbers K_A and K_B such that \delta(A)\le K_A and \delta(B)\le K_B. Take K:=\max\{K_A,K_B\}<\infty (i.e. we take a...
Suppose \delta (A)<\infty. Let x\in A be arbitrary but fixed. Then d(x,y)\le\sup_{(x,y)\in\{x\}\times A}d(x,y)\le\delta (A). Thus setting r:=\delta (A) we find a (closed) ball \bar{B}_r(x):=\{y\in X:d(x,y)\le r\}. It contains all points of A, since for all x\in A and all y\in\bar{B}_r(x) we have...
I am a bit disturbed by the following elementary observation.
Let (X,d) be a metric space and \emptyset\neq A \subseteq X.
(a) The diameter \delta (A) of A is defined to be \delta (A):=\sup_{(x,y) \in A^2}d(x,y), where A^2:=A \times A
(b) A is called bounded if \delta (A)<\infty.
Now let...
Hi!
I wonder how to prove that if y(t)=sin(t) solves an autonomous ODE f(y,y',...,y^(n))=0, then x(t)=cos(t) is also a solution.
I mean I'm a bit distracted by the fact that all derivatives of y are present here. For example in the equation for a pendulum there are just y and y'' and a...
Ok, I'll try to go into detail, because I'd like to write it down formally and clearly.
So you say that:
Constant functions are trivially continuous and holomorphic everywhere in C. U is open, so f' is holomorphic on U. Also the path gamma is continuous and hence smooth. These two statements...
Yes, I think I can.
We have (f\circ{\gamma})'(t)=u'(t)+iv'(t)=0\Longrightarrow{u'(t)=0=v'(t)}, which means that u(t)=C_1, v(t)=C_2 are constant functions and so (f\circ{\gamma})(t) is constant.
Must I use path integrals here. Is the above argumentation incorrect or imprecise?
Please check whether this makes sense
Homework Statement
If U\subset\mathbb{C} open, path-connected and f:U\longrightarrow\mathbb{C} differentiable with f'(z)=0 for all z\in{U}, then f is constant.
Hypotheses:
H1: U is pathconnected
H2: f:U\longrightarrow\mathbb{C}, f'(z)=0, z\in{U}
2. The...
No, the greater-equals-sign is correct.
Estimatation from above is:
|\frac{nx(7+sin(nx))}{4+n^2x^2}|\le{\frac{8n|x|}{4+n^2x^2}, because |sin(nx)|\le{1}...but it leads me to nowhere because I need a majorant independent of x.
Uniform convergence
Hello!
I've got a short question to an example.
I should check the following sequence for uniform convergence on the whole of \mathbb{R}:
f_n(x)=\frac{nx(7+sin(nx))}{4+n^2x^2}
It says that the conevergence is nonuniform, because...