Is this function continuous and differentiable at $x=0$? I'm a new math learner, and this question really concerns me though it's merely a freshman level one.
Let $X\subset \mathbb{R}$, $X=\{0\}\cup(\cup_\text{n is odd}[\frac{1}{n+1},\frac{1}{n}])$, and let $f:X\rightarrow \mathbb{R}$, $f(x):=x$. 
Is this function continuous and differentiable at $x=0$? At least, I think 0 is a limit point to $X$.
Thanks in advance for whoever can help me.
Cheers!~
 A: This is more a question of the details of your definitions have been set up.
Intuitively, I would be comfortable with calling the function continuous and differentiable as a function $X\to\mathbb R$, but probably not as a partial function $\mathbb R\to\mathbb R$.
Further, I would adopt a definition of at least "continuous" that made this true. Continuity at $x_0$ makes sense to me as long as $x_0$ is a point of the domain that's not isolated, and it's easy to say that in a definition, by giving the domain of the function the subspace topology.
For "differentiable", I'm less sure that it would be worth it to figure out a definition that is general enough to match my intuition here.
The overall points here are that:


*

*the details of definitions are a matter of choice, striking a balance between ease of use and being general enough to apply to the case you're interested in,

*not all authors/texts agree on these details.

*if you extend the definitions to cover these cases you need to at least consider whether the results you're using still hold -- by checking whether the source of those results use similarly broad definitions, or alternatively by checking whether you can adapt their proofs nevertheless,

*in most practical cases such results will apply. But that doesn't mean you don't need to check!

