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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 $a<b$, 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?

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  • $\begingroup$ Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals. $\endgroup$ – William Elliot Jul 27 '18 at 10:39
  • $\begingroup$ @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? $\endgroup$ – Saaqib Mahmood Jul 27 '18 at 11:03
  • $\begingroup$ Yes, that the definition is excessive. $\endgroup$ – William Elliot Jul 27 '18 at 20:40
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Your solution is clear and accurate. Well-done!


This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence

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

should instead read

. . . more than one point.

Apart from this, I have no comments for improvement.

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