Recent content by utleysthrow

okay, I guess it wasn't. f IS continuous at (1, 0), and the function value of any path into (1, 0) is always 0. For f to be continuous at (1,0), this condition must hold: lim_{(x, y)\rightarrow(1, 0)} f(x,y) = f(1,0)=0 First, if y=0, then f(x, y)=f(1, 0)=0. Second, if y\neq0, since (x,y)...

Is my example of such a path (that I stated above) not valid? If (xn, yn) = (1, 1/n), it goes to (1,0), but the limit of the function value, lim f(xn, yn) = lim e^{-1/\frac{1}{n^{2}}} = lim e^{-n^{2}}, which does not equal 0... it goes to infinity as n goes to infinity.

Homework Statement f:R2 -> R f(x,y) = e^{-x^{2}/y^{2}} if y is not 0, and 0 if y is 0 a) At (1,1), is f continuous? b) At (1,0), is f continuous? Homework Equations The function f is continuous at the point c if for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c)...

Right. I have kind of a silly question though... how can we assume that xi+yi converges to anything?

And if that is correct, it would follow that L-M is in Y since Y is a closed set. Then M+(L-M)=L is in X+Y

I would say that yj has to converge to L-M... but I'm not sure how to justify that. Does it simply follow from the fact that xj+yj converges to L, and xj converges to M, so (xj+yj)-xj=yj converges to L-M?

If X is compact... there is a subsequence of {xi} that converges to a limit x in X?

If X is compact, the seq {xi} should converge to some x \in X

Now that I think about it... X+Y is still NOT closed even if it contains points other than 1/n, isn't it? That doesn't change the fact that 0 is still a limit point of X+Y, which is not contained in the set. Finding one limit point not in the set is all it takes to make the set NOT closed, no?

Oh, right. I didn't catch that...

X=N Y=\left\{ -n+\frac{1}{n} : n \in N \right\} X+Y=\left\{\frac{1}{n} : n \in N \right\} Thank you for your reply. Closed set, as far as I know, contains all of its limit points. Using that definition, my X+Y is not closed because 0 is a limit point of X+Y, but it is not contained in...

If X=N and Y={-n+1/n} where n is a natural number, then both are closed but the sum X+Y={1/n} is not, since it doesn't include 0 which is a limit. If one of them, say Y, is compact, this doesn't happen because {-n+1/n} has to be bounded by the definition of compactness, which means not all X=N...

Homework Statement I am trying to prove that, if X is compact and Y is closed, X+Y is closed. Both X and Y are sets of real numbers. Homework Equations The Attempt at a Solution I know that a sum of two closed sets isn't necessarily closed. So I presume the key must be the...

Oops, I meant to say a set containing "every" rational numbers, as the question asks. (And yes I'm talking about subsets of R.) As you said, the rationals are dense in the reals, so R=\overline{Q}. So an open set containing every rationals also contains every irrationals, since any open set...

Hmmm... I'm thinking that maybe 6 is true. All rational #s by themselves are closed and so the (countably) finite union of them would be closed... but is it possible to construct an "open" set that just contains rational #s? Edit: actually... regardless of whether it's possible to create an...