If $f$ is continuous and $f(x)$ has no cluster points as $x \to b^-$ then $f(x) \to \infty$ or $f(x) \to -\infty$ as $x \to b^-$ While thinking about another problem, I came up with this somewhat interesting fact:

Suppose $f : (a, b) \to \mathbb{R}$ is continuous, and assume that $f(x)$ has no (finite) cluster points as $x \to b^-$.  Then either $f(x) \to \infty$ as $x \to b^-$, or $f(x) \to -\infty$ as $x \to b^-$.

(Here, the "cluster point" condition specifically means: let $\mathcal{F}$ be the filter on $(a,b)$ associated with $x \to b^-$, i.e. the filter with basis $\{ (b - \epsilon, b) : \epsilon > 0 \wedge a \le b - \epsilon \}$.  Then we are asserting that the filter $f_* \mathcal{F}$ on $\mathbb{R}$ has no cluster point in $\mathbb{R}$.  Unfolding the definitions, this means $L \in \mathbb{R}$ would be a cluster point of $f(x)$ as $x \to b^-$ if for all $\epsilon > 0$ and $\delta > 0$, there exists $x \in (a, b)$ such that $b - x < \delta$ and $|f(x) - L| < \epsilon$; and we are asserting there is no such $L$.)
So, the question would be to give a proof of this fact.
 A: Consider $\limsup_{x \to b^-} f(x)$ and $\liminf_{x \to b^-} f(x)$.  If either of these were finite, then it would be a cluster point of $f(x)$ as $x \to b^-$.  Now, if $\liminf_{x \to b^-} f(x) = +\infty$, then this implies $f(x) \to \infty$ as $x \to b^-$; and similarly, if $\limsup_{x \to b^-} f(x) = -\infty$, then this implies $f(x) \to -\infty$ as $x \to b^-$.  Therefore, the only remaining case to eliminate is where $\liminf_{x\to b^-} f(x) = -\infty$ and $\limsup_{x\to b^-} f(x) = +\infty$.  Now, for any $\delta > 0$, this implies there are $x, y \in (a, b)$ with $b - x < \delta$ and $b - y < \delta$ such that $f(x) > 1$ and $f(y) < -1$.  By the Intermediate Value Theorem, we can find $z$ between $x$ and $y$ such that $f(z) = 0$; and we then know that $z \in (a, b)$ and $b - z < \delta$ also.  This implies that 0 would be a cluster point of $f(x)$ as $x \to b^-$, giving a contradiction.

What I find interesting about this is that the statement is definitively false if we try to make the same statement about sequences: for example, $a_n := (-1)^n n$ has no (finite) cluster points, but $a_n \not\to \infty$ as $n \to \infty$ and also $a_n \not\to -\infty$ as $n \to \infty$.  The statement also fundamentally requires $f$ to be defined for $x < b$ sufficiently close to $b$: otherwise, for example, we could take $f$ to be defined on $\bigcup_{n=1}^\infty [-\frac{1}{2n}, -\frac{1}{2n+1}]$ such that $f(n) = (-1)^n n$ on $[-\frac{1}{2n}, -\frac{1}{2n+1}]$.  So, the fact is fundamentally about the filter $x \to b^-$ and continuous functions on an element of this filter.  (Though the same fact is also true about the filter $x \to \infty$ with basis $\{ (N, \infty) : N \in \mathbb{R} \}$, and similarly about the filters $x \to a^+$ and $x \to -\infty$, with very similar proofs in each case.  It wouldn't be true about the filter of punctured two-sided neighborhoods of $a$, though.)
