Open sets in R^2: Are open sets contained in their closures?

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[SOLVED] open sets in R^2

Homework Statement


If A, B are open in R^2, and A is contained in the closure(B), then is it true that A is contained in B?

Homework Equations


The Attempt at a Solution


It seems intuitively true.

If a is in A, then there is a sequence b_n in B that converges to a by definition of closure. But I don't know what to do with that?
 
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It doesn't seem intuitively true to me. Take B=R^2-{(0,0)}. Give me a counterexample.
 
Last edited:
Dick said:
It doesn't seem intuitively true to me. Take B=R^2-{(0,0)}. Give me a counterexample.

any open set containing the origin
 
Yep.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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