# Prob. 1, Sec. 25, in Munkres' TOPOLOGY, 2nd ed: The components and the path components of $\mathbb{R}$ with lower limit topology

Here is Prob. 1, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:

What are the components and the path components of $$\mathbb{R}_l$$? What are the continuous maps $$f \colon \mathbb{R} \to \mathbb{R}_l$$?

My Attempt:

First of all, we note that $$\mathbb{R}_l$$ denotes the set of real numbers with the topology having as a basis all the open intervals of the form $$(a, b) = \{ \ x \in \mathbb{R} \ \colon \ a < x < b \ \}$$, where $$a$$ and $$b$$ are any real numbers such that $$a, and all the closed-open intervals $$[c, d) = \{ c \} \cup (c, d)$$, where $$c$$ and $$d$$ are any real numbers such that $$c < d$$. Refer to Sec. 13 in Munkres.

Now let $$A$$ be a set in $$\mathbb{R}_l$$ consisting of more than one points. Let $$a, b \in A$$ such that $$a < b$$. Then $$a \in (-\infty, b)$$ and $$b \in [b, +\infty)$$; these two rays are disjoint and their union is all of $$\mathbb{R}$$; moreover these rays are open in $$\mathbb{R}_l$$. So $$A \cap (-\infty, b)$$ and $$A \cap [b, +\infty)$$ is a separation of $$A$$. Thus $$A$$ is not connnected.

Thus we have shown that every set of real numbers having more than one point fails to be connected in $$\mathbb{R}_l$$. In other words, the only nonempty connected sets in $$\mathbb{R}_l$$ are the singleton sets of real numbers.

Since each component of $$\mathbb{R}_l$$ must be connected [Refer to Theorem 25.1 in Munkres.], we can conclude from the preceding paragraph that each component of $$\mathbb{R}_l$$ is a singleton set.

And, since each path component of $$\mathbb{R}_l$$ must be contained in a component [Refer to Theorem 25.5 in Munkres.], we can conclude from the preceding paragraph that each path component of $$\mathbb{R}_l$$ must be a singleton set too.

Finally let the map $$f \colon \mathbb{R} \to \mathbb{R}_l$$ be continuous. Then since $$\mathbb{R}$$ is connected, so should be $$f ( \mathbb{R})$$ [Refer to Theorem 23.5 in Munkres.], and therefore $$f ( \mathbb{R})$$ must be a singleton set in $$\mathbb{R}_l$$. This shows that $$f$$ is constant.

Is this solution correct? If so, is it clear enough too, especially for a beginning student? If not, then where are the issues?

• Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals. – William Elliot Jul 27 '18 at 10:39
• @WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$(a, b) = \bigcup_{x\in(a,b)}[x,b).$$ Is this what you meant to point out? – Saaqib Mahmood Jul 27 '18 at 11:03
• Yes, that the definition is excessive. – William Elliot Jul 27 '18 at 20:40

Now let $A$ be a set in $\mathbb{R}_l$ consisting of more than one points.