MHB Proof that Arcsin x is continuous ....

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Can someone please help me to prove that the function f(x) = Arcsin x is continuous on the interval [-1, 1] ...

Peter
 
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What is the definition of continuous function in general?
 
An invertible function, y= f(x), is continuous at x= x_0 if and only if y= f^{-1}(x) is continuous at x= f(x_0).
 
Cbarker1 said:
What is the definition of continuous function in general?
The definition of continuity in $$\mathbb{R}$$ is given in Stephen Abbott's book: Understanding Analysis, as follows:
View attachment 9348
Alternative characterizations of continuity are given by Abbott in Theorem 4.3.2 as follows:
View attachment 9349So to show (from first principles) that $$\text{Arcsin } x$$ is continuous on $$[-1, 1]$$ we would have to show that given an arbitrary point $$c \in [-1, 1]$$ that for every $$\epsilon \gt 0$$ we can find $$\delta \gt 0$$ such that

$$\mid x - c \mid \lt \delta \ \Longrightarrow \ \mid \text{Arcsin x } - \text{Arcsin } c \mid \lt \epsilon$$ ...But how do we proceed ... ?

Peter

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HallsofIvy said:
An invertible function, y= f(x), is continuous at x= x_0 if and only if y= f^{-1}(x) is continuous at x= f(x_0).
Thanks for the help, HallsofIvy ...

Peter
 

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  • Abbott - Defintion 4.3.1 ... Continuity ... .png
    Abbott - Defintion 4.3.1 ... Continuity ... .png
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  • Abbott - Theorem 4.3.2 ... Characterizations of Continuity ... .png
    Abbott - Theorem 4.3.2 ... Characterizations of Continuity ... .png
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We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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