Recent content by Esd

Considered? Yes. But can't think of how.

"mixed" logarithmic equation Homework Statement I'm trying to solve for x in the following equation: e^(-66/x)/x^2 = c, where c is a constant Homework Equations The Attempt at a Solution By taking ln of both sides and then dividing by 2, I get to: -33/x -ln(x) = c/2 Then, in order to get an...

If by simplest, you mean the singleton set - then no. Because Lim(singleton) is empty.

I was just reading the description of the Cantor set. It seems to fit. All the points are boundary points, as the open intervals are taken out. Since it's a 1/3 each time of the remaining, the distance between any given point to the next iteration of a point created next to it only get's...

I knew that I filled the interior once I did that. However, I can't think of anything that would solve the problem of making 0 a limit point, without breaking anything else. Earlier I was thinking of having 0 and then adding numbers of the form 1/n. n being an integer. Such would make 0 a limit...

Well, for 0 to be in Lim(A),the immediate neighborhood of 0 (at least in one direction) must also be in A.

I'm still hopeless for this problem. Nor do I know of the famous set you're hinting at.

I don't think that anymore. An example of a countable subset of irrationals would be the following: root(2),root(5), root(2)root(5), root(2)root(2)root(5)..: multiplying root(2) by root(5), such that at least one of root(2) or root(5) appears an odd amount of times.

Well, the irrationals have an empty interior and are uncountable. But due to being dense, don't work. I'm thinking it's maybe some subset of the irrationals, one which excludes a countable subset of the irrationals. But I don't think there exists a countable subset of the irrationals.

That's correct. I had a faulty definition for "uncountable" in my head. I drop that claim. I'm still stuck on how to construct such a set A though.

Hurkyl, Thank you for replying. My reasoning for restricting it to be uncountable is as follows. If A was countable, then we have points of A (say p and q), in between which there are no points of A. Thus,for p and q, we would be able to find neighborhoods which wouldn't contain any other...

Homework Statement Give an example of a nonempty set A in R such that A = Bd(A) = Lim(A) = Cl(A). Bd(A) is the boundary of A, Lim(A) is the set of limit points of A, Cl(A) is the closure of A. Homework Equations Bd(A) = A - Int(A), Int(A) is the interior of A Lim(A) = is the set of...

Suppose cv=0, v =/= 0 then take a new scalar from F (call it a) a*(cv)=a*0 (ac)(v)=0 ca(v)=0 c(av)=0 Then c * (the whole vector space spanned by v) = 0 Since this is a vector space, c must be 0 (I forget the name of this property).

Each line describes the previous line (how many of each symbol there were). After 13112221, 1113213211. I can't see there being any characters but 1,2,3 in any of the later lines, and therefore also any character repeating more than 3 times in a line (ex.1 1 1 1). Also, in the long run, "1" to...

Well, a general year here is 365 days. Dividing it into weeks (by 7) would leave a remainder of 1. Where there would be 52 weeks and 1 day. We can assume that those 52 weeks would mean 52 Sundays (now working with an imaginary 364 day-year, just because it doesn't leave the remainder of 1). To...