Unbounded function challenge?

Question

Our teacher challenged us to come up with a function that is definded on R and is unbounded in every neighborhood of x = a for every "a" of R.
Can you help me with this one?

Answer 1

Actually, I don't think scmi's example was what your teacher had in mind, since the function he gave is both not well-defined (what is f(1)? Is it 1, or ∞, since 1=0.99999... ), and also assumes infinite values, and if your teacher allows a function to assume the value ∞, then the function f(x) = ∞ would surely be the simplest example. But I'll assume that your teacher wants a well-defined function assuming only real values.

Let's see, how about:

f(x) = {0 if x is irrational
q if x=p/q, p∈ℤ, q∈ℕ, gcd(p, q) = 1}

In other words, define f to be zero if x is irrational, and the denominator of the unique representation of x in lowest terms if x is rational. This is well-defined, since there is exactly one representation of x in lowest terms if x is rational. Further, it is easy to show that this function is unbounded on every interval.

Answer 2

Well how about
f(x) = sum of all decimal digits of x

Since the reals are infinitely dense, then you can always pick a real with a recurring decimal on any interval, hence f(x) = ∞ in the neighborhood of every real.
So even though f(2.3) = 5 (for example),
f(x) will be unbounded in the neighborhood of x=2.3.

[Pascal's objections are nitpicky, we can easily modify the fn to handle them.
Yes you can come up with the well-known decimal corner case like 1≡0.999 recurring, or e.g. 2.3≡2.299 recurring. So for those cases, just define f(x) as the minimum possible f(x), thus it will use the non-recurring value, hence f(2.3)=5. Trivial.
Secondly, ok yeah f(±∞) is undefined and has no limit.
So let's just define f(±∞) to be 0, already. Gimme a break...]

(I'm sure you can dream up plenty of others, with rationals, transcendentals etc. which are also infinitely dense.
Wonder if any of them are any practical use?)