I am having some troubles understanding the following, any help to me will be greatly appreciated.


1) Let S1 = {x E R^n | f(x)>0 or =0}
Let S2 = {x E R^n | f(x)=0}
Both sets S1 and S2 are "closed"

>>>>>I understand why S1 is closed, but I don't get why S2 is closed, can anyone explain?<<<<<



2) "A set X is a "metric space" if it admits a function d: X x X -> R such that:
(i) d(x,y)>0 or =0
(ii) d(x,y) = 0 iff x=y
(iii) d(x,y) = d(y,x)
(iv) d(x,z) < or = d(x,y) + d(x,z)"

>>>>>Now, I just don't get the "d: X x X -> R" part at all. How can you multiply two sets together? I don't get the notation, can someone please explain?<<<<<



3) "The mapping f: A->B is "one-to-one" if f(x)=f(y) implies x=y, and f is said to map A "onto" B if f(A)=B. A mapping f: A->B is said to be "invertible" if there is another mapping g: B->A such that g(f(x)) = x for all x E A and f(g(y)) = y for all y E B.
===================
The equation g(f(x)) = x can be valid for all x E A only if f is one-to-one, and the equation f(g(y)) = y can be valid for all y E B only if f maps A onto B. Conversely, if these two conditions are satisfied, then f is invertible." <<<<<-These are quoted from my textbook, but I don't understand how they come up with the conclusions in these last two sentences. Could someone kindly explain the reasonings?>>>>>

Thank you!

1. Are you supposed to prove or disprove this for all functions f(x), or were given a specific function? If it is the former, you should easily be able to form a function f(x) for which both statements are false.

2. This is standard notation. The product of two sets is called the Cartesian product (http://en.wikipedia.org/wiki/Cartesian_product). You are already working with Cartesian products. For example \mathcal R^3 is the Cartesian product \mathcal R\times\mathcal R\times\mathcal R. In this case, this is talking about a function d that takes two arguments, both elements of X, and the value of the function is in R. For example, the function that computes Euclidean distance between two points in Rn is one such function (and you should be able to prove that this function satisfies all of the conditions for a distance metric).

3. For the two phrases, "... can be valid ... only if", construct counterexamples (e.g., f is not one-to-one, f does not map A to B). The final phrase (conversely, ...) is almost exactly the definition of invertible.

The binary operations you have used are examples of functions on such domains. In linear algebra, r-linear forms on some set (a module) are a more famous example.

I am guessing 1 refers to some function f from R(n) to R(m) which is continuous on the domain.

1) Sorry, I missed the assumption that f(x) is continuous. Now assuming f(x) is continuous, I don't get why S2 is closed...

3) Is there any difference between "only if" and "if and only if" ? Are they equivalent in meaning?

If, only if, and if and only if (iff) have distinct meanings in math. Truth table:

x y x if y x only if y x iff y
false false true true true
false true false true false
true false true false false
true true true true true


Notes:
"x if y" means that x must be true if y is true. When y is false, "x if y" doesn't say anything about the validity of x. "x if y" is false only in the one case where x is true and y is false.
"x if y", written as x\leftarrow y, is another way of saying "y implies x" (material implication), written as y\rightarrow x
"x only if y" is exactly the same as "y if x" or "x implies y".
"x iff y", written as x\leftrightarrow y, is very strong: it means x and y are equivalent. Another way to write x iff y is x\equiv y


In terms of the OP, read the statement "x can be valid only if y" as "if y is not true then x cannot be true". For example, if f is not one-to-one then g(f(x))=x cannot be valid for all x in A.

1) Sorry, I missed the assumption that f(x) is continuous. Now assuming f(x) is continuous, I don't get why S2 is closed...


The easiest way to see it is probably to consider the complement of S2 (the subset of the domain where f(x) is not equal to 0), and convince yourself that it is an open set.