Confusion with open set and a subset of R2

  • Thread starter Thread starter Heresy
  • Start date Start date
  • Tags Tags
    Confusion Set
Heresy
Messages
54
Reaction score
0

Homework Statement



Using the definition of an open set, prove that the subset S = \left({a}_{1}, {b}_{1}\right)\times \left({a}_{2}, {b}_{2}\right) of the Euclidean space {R}^{2} is open.

Homework Equations





The Attempt at a Solution



I so far don't have much of an attempt, as I am merely trying to figure out what the question is asking. If I interpret the (a1, b1) x (a2, b2) as two vectors on a plane then S would be a vector pointing outwards from the plane? Which doesn't seem like an open set in R2 to me.

I am not looking for a solution, but looking for a clarification of the question in mind.

Thanks!
 
Physics news on Phys.org
(a1, a2) is an open interval on the real number line, just like in high school or calculus.

here, the X refers to the cartesian product. so (a1, b1) X (a2, b2) is the set that consists of tuples1 (x1, x2) where x1 is in (a1, b1) and x2 is in (a2, b2). There is an unfortunate overlap of notation between vector/tuple notation and the notation for intervals.

In symbols and for general sets:

S_{1} \times S_{2} = \left{\{ (a, b) : a \in S_{1}, b \in S_{2} \right}\}1 - tuples are essentially vectors but with no necessarily assumed linear structure
 
Last edited:
Thank you very much!
 
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...
Back
Top