Analytic Characterizations of Continuous Functions of the Lower Limit Topology (2.18.8)

James Munkres Topology, chapter 2.18, exercise 8:

8. (a) Suppose $f:ℝ→ℝ$ is "continuous from the right," that is, $$\lim_{x→a^+}f(x)=f(a)$$ for all $a∈ℝ$. Show that $f$ is continuous when considered as a function from $ℝ_l$ to $ℝ$.
(b) What sort of functions $f:ℝ→ℝ$ are continuous when considered as functions from $ℝ$ to $ℝ_l$? As maps from $ℝ_l$ to $ℝ_l$?

Proof: (a) Let $(a,b)⊆ℝ$. We show that for each $x∈f^{-1}(a,b)$ we have $x∈[x,c)$ for some $c$ so that $f^{-1}(a,b)$ is open. Assume not; then for each $y > x$ we have $f[x,y)⊈(a,b)$. Let $y_1,y_2,...$ be an infinite sequence of terms approaching $x$ from the right such that $f(y_i)∉(a,b)$ for all $i$. But now by hypothesis $f(x)$ is a limit point of $f(y_i)$, implying $x$ is a limit point of the complement of the open $(a,b)$ so that by its closure $f(x)∉(a,b)$, contradiction.

(b) Say $p∈ℝ$ is a local minimum of $f$ if there exists $a,b∈ℝ$ such that $p∈(a,b)$ and $q∈(a,b)⇒f(p)≤f(q)$. We show that $f:ℝ→ℝ_l$ is continuous at $p$ iff $f:ℝ→ℝ$ is continuous and $p$ is a local minimum of $f$, and continue to prove $f$ is continuous iff $f$ is constant. ($⇒$) Let $f:ℝ→ℝ_l$ be continuous at $p$. Then $f'=i∘f:ℝ→ℝ$ is continuous at $p$ seeing as it is the composition of two functions continuous at $p$, the inclusion $i:ℝ_l→ℝ$ being continuous since $ℝ_l$ is finer than $ℝ$. Now, let $(a,b)$ be an open neighborhood of $p$ within $f^{-1}[f(p),∞)$; we have $q∈(a,b)⇒f(q)∈[f(p),f(p)+1)⇒f(p)≤f(q)$ so that $p$ is a local minimum. ($⇐$) Let $f(p)∈[c,d)$. Choose a neighborhood $U$ of $p$ on which $p$ is minimum, and a neighborhood $V$ of $p$ for which $f(V)⊆(-∞,d)$, so that $U∩V$ is a neighborhood of $p$ for which $f(U∩V)⊆[f(p),d)⊆[c,d)$, so $f:ℝ→ℝ_l$ is continuous at $p$.

Suppose $f:ℝ→ℝ_l$ is continuous, so that it is continuous at each point $p$, so that every point of $ℝ$ is a local minimum of $f$ and $f:ℝ→ℝ$ is continuous. Suppose $f(x)≠f(y)$ for some $x < y$, and assume $f(x) > f(y)$ as the argument will be symmetric; let $c=\text{sup }\{z~|~t∈[x,t]⇒f(t)≥f(x)\}$. Evidently if $f(c)≥f(x)$ then $c$ is not a local minimum since for each neighborhood of $c$ we have some element $d > c$ within that neighborhood such that by the supremum definition of $c$ we see $f(d) < f(x) ≤ f(c)$. But if $f(c) < f(x)$ by some margin $ε$ we observe each neighborhood of $c$ contains some $d < c$ such that $f(d)≥f(x)$ and hence $f(c) < f(d)$ by a margin at least as large as $ε$ so that $f:ℝ→ℝ$ cannot be continuous at $c$.

Similarly, we can show that a map $f:ℝ_l→ℝ_l$ is continuous at a point $p$ iff $f:ℝ→ℝ$ is continuous at $p$ and $f$ is locally increasing at $p$, that is, there is $ε > 0$ such that $f$ is increasing on $[p,p+ε)$. By a similar method we find that the continuous functions are precisely the increasing functions.$~\square$