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

GRE question:

Above is from here:

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

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,

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

• Yes, what you say is right. The solution you quoted is sloppy. Oct 24, 2016 at 5:03
• @symplectomorphic Thanks ^-^ Sloppy or not, it's more accessible to me than the other solutions. Post as answer?
– BCLC
Oct 24, 2016 at 5:05
• I don't think this question really deserves to be a question, so I'm not posting an answer: this is just a matter of notation, and you already answered your own question. Doesn't seem right for me to post something when a simple "yes, you're right" is all that's required. Oct 24, 2016 at 5:10
• yes, it is sloppy; that doesn't mean it isn't just a matter of notation. of course $g(0)$ need not be defined, for we are only given a function on $(0,1)$. so of course you are right, strictly speaking, but the idea is clear: the existence of a continuous extension requires that the desired limit exist. Oct 24, 2016 at 5:36

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 $1$.
• $g(0)$ and not $\tilde{g}(0)$ ?
• Whatever you like. You can call the extension $\tilde{g}$ or $g$ depending on your pedantic mood. Oct 24, 2016 at 5:08