Recent content by Congruent

Yes. The argument is just a definition chase. Let (x,y) be a point in the product X x Y, and fix an r > 0. We need to find a point (p,q) in P x Q so that (p,q) lies in the ball B( (x,y); r). Since P is dense in X, we may find a point p in P so that p is in B(x;r). Similarly, find a q in Q so...

It is a standard fact that if f is a real valued function defined on a closed interval which is integrable in the Riemann sense, then the function is measurable and integrable (in the Lebesgue sense) and the two integrals are equal. So if what you're asking for is an example of finding an...

I see. But I'm only required to make countably many such choices, no? So wouldn't this still only be an instance of countable choice? Sorry, I'm entirely self-taught, and most of these things are discussed without any mention of choice being used. I suppose mostly because it's so...

Really? I imagine it has to come up in if for every sequence \{x_n\}_{n=1}^{\infty} which converges to x, it follows that f(x_n) \to f(x), then f is continuous at x, seeing as the other direction is pretty straight forward. But I imagine to prove this, you'd do something like suppose f is not...

I can't come up with a single theorem appropriate to discuss in a first calculus course that requires using Zorn's lemma or any transfinite method. Could you provide an example?

Yes. Using the definition above, let \{x_{k_n}\}_{n=1}^{\infty} be any convergent subsequence. By assumption, there is an M > 0 so that x_k \le M . In particular, M \ge x_{k_n}. Thus M \ge \lim_{n \to \infty} x_{k_n} . This shows that M is an upper bound to any subsequential limit...

Possibly I'm having trouble with the formatting, but it looks an awful lot like you used the closed brackets. In any event, apologies for the misunderstanding.

Not necessarily. The question posed by the OP is if he can find a metric turning (-\infty, 5] into a metric space, then is (4,5] necessarily open. This is the question as it's posed, and the answer is not necessarily. He did not specify what the metric would be, so there's no guarantee that...

I disagree. Homeomorphisms preserve, among other things, compactness. While [-\pi/2,\pi/2] is compact (by, say, Heine-Borel), \mathbb{R} is not.