Give an example to a function $ f: (a,b) \to \mathbb R $ where $ f_x $ is not bounded, but every point $x \in (a, b) $ has a neighborhood where the function is bounded.

(If I translated neighborhood wrong, please correct me)

I gave:

$ f_x = \frac 1{x - b} $

Because the only point having no neighborhood where $ f_x $ is bounded is $ x = b $, and $ b \notin(a,b) $

Please verify

Update: I added the proof on which I relied.

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    $\begingroup$ Its fine. Any function which "blows" up only at infinity will do. $\endgroup$ – Samrat Mukhopadhyay Feb 12 '15 at 18:28
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    $\begingroup$ Well not really since the function is only defined on $(a,b)$. $\endgroup$ – Tim Raczkowski Feb 12 '15 at 18:33
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    $\begingroup$ Your example is correct Dean. $\endgroup$ – Tim Raczkowski Feb 12 '15 at 18:34
  • $\begingroup$ @TimRaczkowski: Sorry, you're right, I clarified that it's undefined in b. Thank you. $\endgroup$ – Dean Feb 12 '15 at 18:37
  • $\begingroup$ it is correct :) $\endgroup$ – Anubhav Mukherjee Feb 12 '15 at 18:46

I will write the proof so that my question contributes (the question is a sequel to another question in the homework and my proof relies on it).

We will demonstrate this by looking at $f:[a,b] \to \mathbb R$ that is not bounded and showing that for $f:[a,b]$ exists a point $x_0 \in [a,b]$ that has no neighborhood in which $f:[a,b]$ is bounded. Then we write $f_x$ for which this point is either $x_0 = a$ or $x_0 = b$, and since we are asked to give an example for $f:(a,b) \to \mathbb R $ the $f_x$ we write will be correct.

Let $f:[a,b] \to \mathbb R$ not bounded in the interval $[a,b]$, therefore for any $M$ exists $x_0 \in [a,b]$ so that $f_{x_0} \gt M$, in particular for any $M \in \mathbb N$.

We can build a sequence of $x_n$ so that for any $M$ exists $x_n$ so that $f_{x_n} \gt n = [M] + 1$, and since $f_x$ is not bounded in the interval there is an infinite amount of $M$'s and $x_n$'s in the interval.

The interval $[a,b]$ is bounded by $a$ and $b$ therefore from Bolzano–Weierstrass theorem we infer that the sequence $x_n$ has a sub-sequence $x_{n_k}$ which converges to a final limit $x_{n_k} \to x_0$. The sequence is bounded $a \le x_{n_k} \le b$ therefore $x_0 \in [a,b]$.

$x_{n_k} \to x_0$, therefore for any $\varepsilon$ exists $N$ so that for any $n \gt N$ we have $|x_{n_k} - x_0| < \varepsilon$. Mark $N = [M] + 1$ and we have that for any $M$ there is $n_k \gt [M] + 1$ for which have a neighborhood where any $x_{n_k} \in[x_0 - \varepsilon, x_0 + \varepsilon]$ is $x_{n_k} \gt [M] + 1$.

We showed that for an unbounded function $f:[a,b] \to \mathbb R$ exists a point $x_0 \in [a,b]$ for which $f_x$ is not bounded in any of its neighborhoods. For short, it's the point where the function has an asymptotic discontinuity (or perhaps something else I am not familiar with yet).

Now, back to the example, we write a function that has an asymptote at a single point that is a limit point of it's set. From the above proof, while this point is not in the set, it will have no neighborhood in the set where the function is bounded.

So for $f:(a,b) \to \mathbb R$ that is $f_x = \frac 1{x-b}$ we have an asymptote at $x \to b$ thus $f_x$ is not bounded in any neighborhood of $b$ but $b\notin(a,b)$.

Hope there's nothing I missed.

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