Doing a problem on rings from Dummit & Foote I think I'm mis-reading it

[farleyknight]

Homework Statement

Decide which of the following are subrings of the ring of all functions from the closed interval [0,1] to R (the reals)

a) The set of all functions f(x) such that f(q) = 0 for all q in Q (the rationals) & q in [0, 1]
b) The set of all polynomial functions
c) The set of all functions which have only a finite number of zeros, together with the zero function
d) The set of all functions which have an infinite number of zeros.
e) The set of all functions f such that lim {x -> 1+} f(x) = 0
f) The set of all rational linear combinations of the functions sin(nx) and cos(mx) where m and n are non-negative integers

Homework Equations

The Attempt at a Solution

The first one is pretty straight forward to show that it is a subring. But unless I'm mistaken, the sets mentioned from b to f aren't even subsets let alone subgroups or subrings since they can be defined on a larger domain than [0,1]. Am I correct? Or am I reading it wrong?

 

[jbunniii]

The first one is pretty straight forward to show that it is a subring. But unless I'm mistaken, the sets mentioned from b to f aren't even subsets let alone subgroups or subrings since they can be defined on a larger domain than [0,1]. Am I correct? Or am I reading it wrong?

I'm sure they mean that you should assume in all six parts that the functions are from [0,1] to $\mathbb{R}$, except that doesn't make sense for part (e). Does it really say

$$\lim_{x \rightarrow 1^+}f(x)$$?

Because that limit doesn't make sense if the function's domain is [0,1].

[Edit]: I just checked using the "look inside" feature on Amazon; in that edition (3rd) it reads

$$\lim_{x \rightarrow 1^-}f(x)$$

which makes more sense.

 

[farleyknight]

I'm sure they mean that you should assume in all six parts that the functions are from [0,1] to $\mathbb{R}$, except that doesn't make sense for part (e). Does it really say

$$\lim_{x \rightarrow 1+}f(x)$$?

Because that limit doesn't make sense if the function's domain is [0,1].

I'll take your word for it since that's the only conclusion I could come to myself.

But as for part (e), that is correct. You can search for the question inside the only online version I could find (page 231 in that edition):

https://www.amazon.com/dp/0471433349/?tag=pfamazon01-20

 

[jbunniii]

I'll take your word for it since that's the only conclusion I could come to myself.

But as for part (e), that is correct. You can search for the question inside the only online version I could find (page 231 in that edition):

https://www.amazon.com/dp/0471433349/?tag=pfamazon01-20

Ha, you read my mind (see my edit above)! But I'm looking at 6(e) on page 231 in the Amazon viewer right now, even zooming into make sure. That's a minus sign, not a plus sign.

 

[farleyknight]

Ha, you read my mind (see my edit above)! But I'm looking at 6(e) on page 231 in the Amazon viewer right now, even zooming into make sure. That's a minus sign, not a plus sign.

Ah crap, you're right. I read it wrong. Well, in my edition it is + not -.. I should probably get a newer edition :(

 

[jbunniii]

Ah crap, you're right. I read it wrong. Well, in my edition it is + not -.. I should probably get a newer edition :(

Yeah, I have an older edition too, I think the 2nd (I don't have it here with me). It has its share of typos but I doubt it's worth spending the money to upgrade unless you really want the new material from the 3rd. There is an errata list available here, in case you don't already have it:

www.emba.uvm.edu/~foote/errata3w.pdf