Is this function continuous and differentiable at $x=0$?

Question

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!~

Answer 1

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!