Closed Sets in XxY & Recognizing W in RxR

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Homework Statement


1. If A and B are closed. Show that A x B is a closed subset of X x Y with the product topology.

2. Give an example of a closed subset W of RxR such that the first component of W (it is made up of two components) is not closed in R.

For 2. How does one recognise a closed subset W in RxR?

For 1. How do you show A x B is closed in general? Is the only way from functions like pi and when one end is open or closed the other must be as well.
 
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RxR is R^2, the plane, and closed subsets are just sets containing their boundaries. For example, points, lines, line segments (with endopints included), closed disks (ie, including the boundary circle), and any finite union of these are closed.

What is your definition of the product topology? One definition makes this problem less than trivial. For the other, which is probably the one you're using, you need to show that the complement of AxB is open, ie, is a union of basis sets. This can be done by some set theory, made easier by drawing a picture.
 
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I've done 1) by using the pi mapping and using the property of cty of pi.

What about an example for 2)?
 
I'll give you a hint: the graph of a continuous function f(x) is closed (prove this). Try looking at some familiar functions.
 
How about the example W={x,y|y=1/x} which is closed in R^2 because W contains all its adherent points.

The first component of W is R/{0} which is not closed as 0 an adherent point of R but not in W.

Correct?
 
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Yea, that works.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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