Recent content by sonofagun

Ah, I see. It wouldn't make sense to say f is continuous or discontinuous at a point outside its domain because f isn't defined (doesn't exist) there. Thanks for clearing that up. I'm reading an Analysis book, by the way. It also contains a topological definition which I haven't gotten to yet.

I haven't covered limit points yet. I'll keep this in mind though.

Maybe I am confused. The definition of continuity says that if f is continuous at x, then f must be defined at x. Hence, I assumed that if f is not defined at x, then f is not continuous at x. For example, the function f(x)=1/x-2 is not defined at 2. Doesn't this imply that f is not continuous at 2?

Thanks for your help. Regarding the points x1...xn: The definition says at most finitely many points can be removed from the domain. It may be that no points are missing from the domain, in which case the domain is ℝ. But, if there is a set points not in the domain, it can only be finite...

Suppose at least one of limz→x-f(x) or limz→x+f(x) does not exist. Then limz→xf(z) does not exist, and therefore ≠ f(x). Hence f is discontinuous at x. Suppose there exists a δ>0 such that f is bounded on {z∈D : distance (z, x) <δ}. If there are sequences zn and wn that converge to x (where zn...

Let D⊆ℝ be an interval of nonzero length from which at most finitely man points x1,...,xn have been removed and let f: D→ℝ be a function. Then every discontinuity x∈D∪{x1,...,xn} of f is one of four types (removable, jump, infinite or discontinuity by oscillation). Proof: Let x∈D or let x be...

I was using the colloquial meaning of necessary.

New topics always involve new notation/symbols, or a rehashing of familiar notation/symbols. Quite often you will have to stop and think about what it means, and translate it to plain English before reading further. But, like the poster above said, it becomes visual given enough study. A vague...

I guess what I'm struggling to understand is why the existence of the limit of f(g(x)) depends on these assumptions. For example, if limy→L f(y) did not exist or if g[I\{x}] was not contained in J\{g(x)} would that imply that the limit of f(g(x)) doesn't exist? If so, why?

I need help with the following theorem: Let I, J ⊆ℝ be open intervals, let x∈I, let g: I\{x}→ℝ and f: J→ℝ be functions with g[I\{x}]⊆J and Limz→xg(x)=L∈J. Assume that limy→L f(y) exists and that g[I\{x}]⊆J\{g(x)},or, in case g(x)∈g[I\{x}] that limy→L f(y)=f(L). Then f(g(x)) converges at x, and...

Okay, I get it now. It's the set of all reals that are ≥ an for infinitely many n's. aN+1 is an element in this set, thus it's not empty. I'm studying independently so I occasionally get stuck trying to figure out easy concepts like this. Thank you!

I'm reading the proof that a cauchy sequence is convergent. Let an be a cauchy sequence and let ε=1. Then ∃N∈ℕ such that for all m, n≥N we have an-am<1. Hence, for all n≥N we have an-aN<1 which implies an<aN+1. Therefore, the set {n∈ℕ: an≤aN+1} is infinite and thus {x∈ℝ : {n∈ℕ: an≤x} is...

This was the part I was struggling with. I understand the reasoning now. Thank you! I'm reading Mathematical Analysis: A Concise Introduction by Schroder. This proof is exceptionally ugly, and I assume there will be some more ugly ones along the way, but overall it's a decent book.

The book uses the axiom of least upper bounds, not the intermediate value theorem. Let R = {x∈ℝ: x≥0 and xn≤a}. Then, 0∈R and R is bounded above by max {1, a} (verifying this is tedious, but it's true). Since R is bounded above, it has a supremum. Let r=Sup(R). To show rn=a, we will show that...

Some background information: Given a nonnegative real number a, I need to show that there exists a nonnegative real number r such that rn=a. The proof involves showing that rn is not less than a, and not greater than a. If rn<a, then rn<rn+δ<a for some δ. The author then writes the...