GRE 1268 #57 - Does $g(0)$ exist if $g$ is uniformly continuous on $(0,1)$?

Question

GRE question:

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An answer:

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Above is from here:

http://www.math.ucla.edu/~iacoley/gre/Practice%201%20solutions.pdf

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Other answers:

https://drive.google.com/file/d/0B-uVGGkZosoPcGVRcDNPN0d2ZVU/view https://drive.google.com/file/d/0B4qQg_AuKUglUnd6YWIyYWNudWM/view http://www.rambotutoring.com/GR1268-solutions.pdf https://www.youtube.com/watch?v=2soO0qqR4w8

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My question:

I like the extension approach sooooo...

Does $g(0)$ necessarily exist?

I think it's supposed to be that $g$, uniformly continuous on $(0,1)$ extends to $\tilde{g}$, continuous on $[0,1]$. Hence,

$$\lim g(x_n) = \lim \tilde{g}(x_n) = \tilde{g}(\lim x_n) = \tilde{g}(0)$$

Answer 1

If $g$ is uniformly continuous then it maps Cauchy sequences to Cauchy sequences.

Pick Cauchy sequence (in $(0,1)$) that converges to $0$ and use this to define $g(0)$.

Now show that the result is independent of the Cauchy sequence sequence used. It follows from this that the resulting function is continuous at $x=0$.

Repeat for $1$.