This makes sense. Each sequence can be paired with the decimal expansion of an irrational number. The set of irrational numbers is uncountable. Therefore the set of all possible sequences (including infinite sequences), consisting of 0,1,...,9 cannot be contained in an infinite and random...
First off, thanks for the thorough response. I have some follow up questions
How do we reconcile the statement that an event has the possibility of occurring but its probability is 0? Or an event has a probability 1 of occurring but is not guaranteed to occur? I know I was taught to believe...
Instead of allowing the sequence to be constructed from the set of all natural numbers, let's restrict the available terms to 0, 1, ..., 9. Each with a probability of 1/10 of being the next term in the sequence. Is it required that an infinite sequence constructed randomly from these numbers...
Hi,
True or False: Every infinite sequence of natural numbers, who's terms are randomly ordered, must contain every possible subsequence of any length, including infinity.
For example, does the infinite and random sequence \small M of natural numbers require that the subsequence {59,1,6}...
So what if we take the extended real line (which is compact) as the set X. And define the function f(x) = x, for all x in X. Then is the max of f ∞? If so does this not work for the real line? And is f bounded by -∞ and ∞?
This is only true of R^{k}, not metric spaces in general. (The...
Hi
I'm trying to determine the significance of a compact subset of a metric space in relation to calculus in general.
I know the definition: A subset \small K of a metric space \small X is said to be compact if every open cover of \small K contains a finite subcover.
But what is the...
I'm actually wrong. The statement - all interior points are limit points - is not true in general, namely when E is a discrete set it is not true! If E = [0,1] then yes every interior point is a limit point.
proof: if p is an interior point then there exists a neighborhood N s.t. N is a subset...
Claim: isolated points cannot be interior points of any discrete subset E of a metric space
proof: by definition, all interior points are limit points. If p is an isolated point of E, we can find a nbhd N(p) such that p is the only point in N(p). Therefore p cannot be a limit point of E and...
Rudin states "E is open if every point of E is an interior point of E", so if this "if" is in fact IF AND ONLY IF, then a discrete set cannot be open b/c all points in E are isolated.
yes, my mistake, p \in E
Here's how: every point in E is an isolated point => E has no limit points => E contains all of its limit points => E closed. (As far as I can tell, this is true for a discrete set containing finitely many or infinitely many points)
I was wrong to say not open...
Hi,
If X \LARGE is a metric space and E \subset X is a discrete set then is E \LARGE open or closed or both?
Here's my understanding:
E \LARGE is closed relative to X \LARGE.
proof: If p \subset E then by definition p \LARGE is an isolated point of E \LARGE, which implies that p...
I think the issue was that I had assumed bounded sequences converge. This is clearly not the case.
To answer my question using the example you gave, there are 2 subsequences of p(n) which converge: p(2n) and p(2n+1), but a number of other subsequences (infinitely many to be exact) that...
Hi,
In Baby Rudin, Thm 3.6 states that If p(n) is a sequence in a compact metric space X, then some subsequence of p(n) converges to a point in X.
Why is it not the case that every subsequence of p(n) converges to a point in X? I would think a compact set would contain every sequence...