Recent content by mathgirl313

Over a map f that is continuous, and I believe it also has to be onto. I keep trying to come at this at different angles and can't seem to get anywhere. So more formally... If f:X→Y, and f is continuous and onto. Let A be a dense set in Y. Then f^1(A) is dense in X. I'm not entirely...

Thanks for the help! I pretty much get all of it now. The last part though, I know the square root had to be greater than or equal to zero...but when I tried solving I got a different interval. I had 4/3 and 16/3 as coefficients for alpha and beta...but that was used for something different...

So I have the equations x'(t) = 4*y(t) and y'(t)=-x(t). Going through all the steps, you get x(t) = c1*2*cos(2t) + c2*2*sin(2t) and y(t) = -c1*sin(2t) + c2*cos(2t). And of course, this can just can be written as the vector. So after all that you are to find the maximum speed of the...

They both help, big time! Thank you! My biggest problem was that I've not seen a straight up example, just was told this fact, so was having trouble constructing and trying to include the metrics, and figuring out completeness since we didn't do a whole with that either..just all confused me

It's sufficient to say I'm lost...I think I'm exhausted from looking at top. all day.. So we have a complete metric space and an incomplete metric space, they're both on the open interval from -pi/2 to pi/2 into the reals. So all you need are the continuous functions linking the two metrics...

And just to help my confusion, in your post explaining each metric, you said the first metric was cauchy and was not cauchy

The definition I was given for complete is that a space is complete if every Cauchy sequence is convergent in the space. So for the first metric, how could the space couldn't be complete if the sequence was not Cauchy.. The absolute value of arctan should get closer and the functions does not...

I must say I'm a little confused my topology isn't great, so I'm trying to get there. The first metric would be Cauchy and complete I think? The second one wouldn't? Can you use the original functions to to create the homeomorphism? And weren't we trying to construct the homeomorphism between...

f(pi/2) = arctan(pi/2) which is like 1.000?? something isn't it? Do you mean f-1? because tan(pi/2) would be undefined sadly... Making it an open interval wouldn't help would it? Because then an wouldn't be convergent since pi/2 is not in the set.

I'm well aware and understand that homeomorphism do not need to preserve metric completeness, I'm just trying to work out a simple counterexample. I have tried searching around just for kicks, but only seem to find more complex ones. I'm wondering if the one I have works for it for sure? On...

Then 3a - a^2 = 0, and 3Z has no zero divisors, so factoring the equation gives a=0 or a=3. Then, if σ(3)=a=0, then σ(x)=0 for all x in 3Z, and σ is the zero homomorphism. If σ(3)=a=3, then σ(x)=x for all x in 3Z, and σ is the identity homomorphism. Hence there there two ring homomorphism for...

So 3a=a^2 if a=σ(3). So the possible homomorphisms of a will have that property. And it should them just fail out like the other examples.

Homework Statement Find all ring homomorphisms from 3Z to Z, where 3Z are the integers that are of multiple 3. Homework Equations The Attempt at a Solution So 3Z is cyclic so σ(3) is sufficient to look. Now all of the other examples have finite groups, so |σ(a)| divides the |a|...

Awesome! Thank you do much for walking me through it! :smile:

Why is it normal? I can only think that since P is a Sylow P-Subgroup of H, and P is normal in H. But a p-Sylow group is normal if and only if the number of Sylow p-subgroups is one (for that ). So P is the unique Sylow p-subgroup. So P is the unique Sylow p-subgroup in G as well, and as such...