I see what you mean, actually, that's what I thought concerning ##x_kj+c \in X_j## for ##c \in N##, ie ##x_k100+c \in X_100##.
I think I can write a coherent proof now that you've helped me resolve my questions. Thank you very much for having the patience to help me! I appreciate it.
Hmm... I am really pulling a blank here.
I know the general definition of a convergent sequence is that ## \exists k \in N##, such that ##d(x_n,L)< ε##, ##\forall ε > 0##, ##k \geq n##.
Maybe construct a sequence ##x_kn## where ##d(x_kn,L)<ε##, for ##ε>0##? Would this just be the terms from...
Hmm... would it be because if ##L \not\in X_100##, then that would imply that the ball ##B(L,ε)## would not be a subset of ##X_100##, though we've established that when ##n \geq k##, ##\exists k \in N##, ##x_k \in B(L,ε) \subseteq X_(n \geq k)## for ##\forall ε > 0##, implying ##X_(n \geq k)...
Sorry:
Create a sequence ##{x_n}## where ##x_n## ##\in## ##X_n##.
The sequence ##{x_n} \subseteq {X_1}##.
Since the sequence of ##{X_n}## is compact, ##\forall ε >0##, ##\exists x_k##, ##k \in N## such that ##d(x_k, L) < ε ## for ##k \geq n##, and where ##L## is the limit point of the...
Oh, so what you're doing is creating a subsequence of points from each ##X_n##, and saying that this sequence has a limit ##L##?
In this case, shouldn't the limit lie in ##X_1##, and since it's a limit point for a sequence containing points from each ##X_n+1## it should get arbitrarily close to...
When you say ##x_n##, first as a point and then a sequence, do you mean the single point sequence ##{x_n}##?
Shouldn't ##L## be in ##X_n##, implying ##L## is in every ##X_n## before it, up until ##X_1##?
Homework Statement
Let ##{X_n}## be a sequence of nonempty compact subsets of the metric space Ω such that ##{X_n+1}## ##\subseteq## ##{X_n}##, with ##n: 1→∞##. Prove that their intersection is nonempty.
**By the way, I mean to subscript "n+1", not have ##{X_n}## + 1 as it seems...