Prob. 7 (b), Sec. 18, in Munkres' TOPOLOGY, 2nd ed: Continuous mappings between $\mathbb{R}$ and $\mathbb{R}_l$

Question

Let $\mathbb{R}$ denote the set of all real numbers with the topology having as a basis all open intervals $$(a,b) \colon= \{ \ x \in \mathbb{R} \ \colon \ a < x< b \ \},$$ where $a$ and $b$ are any real numbers such that $a<b$.

Let $\mathbb{R}_l$ denote the set of real numbers with the topology having as a basis all the half-open intervals of the form $$[a,b) \colon= \{ \ x\in \mathbb{R} \ \colon \ a \leq x < b \ \},$$ where $a$ and $b$ are any two real numbers such that $a<b$.

Now what functions $f \colon \mathbb{R} \to \mathbb{R}$ are continuous when considered as maps from $\mathbb{R}$ to $\mathbb{R}_l$?

My work:

Let $[a, b)$ be a basis element for the topology of $\mathbb{R}_l$. Now, in order for $f$ to be continuous as a map from $\mathbb{R} \to \mathbb{R}_l$, the inverse image $f^{-1}\left([a,b)\right)$ must be open in $\mathbb{R}$ with the standard topology. If the latter set is empty, then it is (vacuously) open; so suppose $x_0 \in f^{-1}\left([a,b)\right)$. Then $f(x_0) \in [a,b)$; that is $a \leq f(x_0) < b$.

Since $x_0 \in f^{-1}\left([a,b)\right)$ and since $f^{-1}\left([a,b)\right)$ is open, there exists a basis element $(c,d)$ for the standard topology on $\mathbb{R}$ such that $x_0 \in (c,d) \subset f^{-1}\left([a,b)\right)$.

Let $\delta$ be the smaller of the two numbers $x_0-c$ and $d-x_0$. Then $\delta > 0$ and $$(x_0-\delta, x_0+\delta) \subset (c,d) \subset f^{-1}\left([a,b)\right);$$ so $$f\left( (x_0-\delta, x_0+\delta) \right) \subset f\left( f^{-1}\left( [a,b) \right) \right) \subset [a,b);$$ hence $$ f\left( (x_0-\delta, x_0+\delta) \right) \subset [a,b).$$

So for any $x_0 \in \mathbb{R}$ and for any $\epsilon > 0$, there is a $\delta > 0$ such that $$f(x_0) \leq f(x) < f(x_0) +\epsilon$$ for all $x\in \mathbb{R}$ such that $\vert x - x_0 \vert < \delta$.

Am I right? Is the above reasoning correct? Is what I've ended up with the complete answer?

What functions $f \colon \mathbb{R} \to \mathbb{R}$ are continuous when considered as maps from $\mathbb{R}_l \to \mathbb{R}$?

Answer 1

A continuous function $f$ maps connected sets to connected sets. But $\mathbb{R}_\ell$ is totally disconnected, so $f$ is constant.

Answer 2

Continuous functions $\mathbb{R}\to\mathbb{R}_l$:

If the continuous function takes some value $f(x_0)$ then all the values that $f$ take are not smaller than that number.

In fact, you have already shown it locally: The values $f$ takes in a neighborhood of $x_0$ are larger than $f(x_0)$, and arbitraily close to it.

Let $U=f^{-1}([f(x_0),\infty)$. Let $y_0$ be a boundary point of $U$. If $f(y_0)<f(x_0)$ then points near $y_0$ and inside the $U$ would be $\geq f(x_0)$ but arbitrarily close to $f(y_0)$. Therefore $f(y_0)\geq f(x_0)$, and together with it a neighborhood of $y_0$ would also be sent to $[f(x_0),\infty)$. Therefore $U$ can't have boundary, i.e. $U=\mathbb{R}$.

If the function takes two values $a$ and $b$ then all its values are $\geq a$ and $\geq b$. Therefore, $a=b$.

Hence the function must be constant.

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Continuous functions $\mathbb{R}_l\to\mathbb{R}$:

Since for every open $U\ni f(x_0)$ one must be able to find $[a,b)\ni x_0$ such that $f([a,b))\subset U$, then $f$ must be right-continuous (the usual limit from the right is equal to the value at the point).

It is clear that, conversely, all functions that are right continuous are continuous for this choice of topologies.

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Continuous function $\mathbb{R}_l\to\mathbb{R}_l$:

We have already determined the continuous functions $\mathbb{R}_l\to\mathbb{R}$. Since $\mathbb{R}_l$ has a finer topology, then the continuous functions $\mathbb{R}_l\to\mathbb{R}_l$ are a subset of these (the functions that are right-continuous). You can check directly that all of them are still continuous:

At each point $f(x_0)$, if $[a,b)\ni f(x_0)$ there is an open $(c,d)\ni x_0$ such that $f((c,d))\subset [a,b)$. In particular $f([x_0,d))\subset [a,b)$.