Recent content by zacharyh

(=>) Assume f(X \cap Y) = f(X) \cap f(Y). And, assume f is not one-to-one. Then, \exists c \neq d such that f(c) = f(d) = z. Now, let X = \{c\}, Y = \{d\}. Then, f(X \cap Y) = \emptyset \neq f(X) \cap f(Y) = \{z\}. So our assumption that f was not one-to-one is false. Therefore, f must be...

Following from your reply. Let X = \{c\}, Y = \{d\}. Then, f(X \cap Y) = \emptyset \neq f(X) \cap f(Y) ... was it that easy?

Attempt #2 (<=) And yes, I'm restating the "given" information (the assumptions) for clarity. (<=) Assume f is one-to-one. Let b \in f(X) \cap f(Y). Then, b \in f(X), b \in f(Y). Then, there exists c \in X, d \in Y such that f(c) = b = f(d). Since f is one-to-one, c = d. Then, c, d...

Homework Statement General Case: Let f:A \rightarrow B be a function. Show that f( \bigcap_{\alpha\in\Lambda} T_{\alpha}) = \bigcap_{\alpha\in\Lambda} f(T_{\alpha}) for all choices of \{T_{\alpha}\}_{\alpha\in\Lambda if and only if f is one-to-one. Simpler Case: Let f:A \rightarrow B...

Yes I'm aware that the limit is sandwiched between two 0s and is therefore 0. I guess what I am asking for is a proof of the inequality. How did you decide to pick sin(1/n)? How do I know that it is between those functions?

Oh I see...sammich' theorem! How do you produce that inequality? I'm not sure how to work with sin(1/x)... can it be related to sin(x) somehow?

Asked to compute: \lim_{n\to\infty} (-1)^nsin(1/n) I've broken this limit down into: \lim_{n\to\infty} (-1)^n * \lim_{n\to\infty}sin(1/n) I've determined \lim_{n\to\infty}sin(1/n) = 0 Now I have \lim_{n\to\infty} (-1)^n * 0 This is where I run into trouble...