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Simple topology problem involving continuity

Example 1 in James Munkres' book, Topology (2nd Edition) reads as follows:

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Munkres states that the map p is 'readily seen' to be surjective, continuous and closed.

My problem is with showing (rigorously) that it is indeed true that the map p is continuous and closed.

Regarding the continuity of a function Munkres says the following on page 102:

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Let $\displaystyle X$ and $\displaystyle Y$ be topological spaces. A function $\displaystyle f \ : \ X \to Y $ is said to be continuous if for each open subset V of Y, the set $\displaystyle f^{-1} (V) $ is an open subset of X.

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Yes ... fine ... but how do we use such a definition to prove or demonstrate the continuity of p in the example?

Can someone show me how we use the definition (or some theorems) in practice to demonstrate/ensure continuity?

I have a similar issue with showing p to be a closed map.

On page 137 Munkres writes the following:

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A map $\displaystyle p$ "is said to be a closed map if for each closed set $\displaystyle A$ of $\displaystyle X$ the set $\displaystyle p(A)$ is closed in $\displaystyle Y$"

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Again, I understand the definition, I think, but how do we use it to indeed demonstrate/prove the closed nature of the particular map p in Munkres example?

Hope someone can help clarify the above issues?

Peter

NOTE: as an aside, if anyone can tell me how to get images to size to the size of the panel and hence be readable, I would appreciate it.

Re: Simple topology problem involving continuity

Let $U \subset [0,2]$ be open. Let $x \in p^{-1}(U)$ be any point. Since $U$ is open in $[0,2]$, there exists $r>0$ such that $B_r(p(x)) \subset U$. Thus, $B_r(x) \subset p^{-1}(U)$ (If you really want, this can be proven more rigorously by sending $B_r(x)$ through $p$ and showing the image is a subset of $U$). Hence, $p$ is continuous.

To show that it is closed, let $F \subset [0,1]\cup [2,3]$ be a closed set. Let $y \in [0,2]\setminus \left(p(F)\cup \{1\}\right)$ (I am excluding 1 to avoid multiple preimages, as that is the only point with multiple preimages. I will handle that case next). Since $F$ is closed in $[0,1]\cup [2,3]$, there exists $r_0>0$ such that $B_r(p^{-1}(y)) \subset [0,1]\cup [2,3] \setminus F$. Let $r = \min\left\{r_0,\dfrac{|1-y|}{2}\right\}$. Thus, $B_r(y) \subset [0,2]\setminus p(F)$. (Again, if you really want, you can send $B_r(p^{-1}(y))$ through $p$ and show that you get $B_r(y)$).

If $1 \in [0,2]\setminus p(F)$, then you need to handle that separately. Since $p^{-1}(1) = \{1,2\}$, you need to find two radii. Choose $r_1>0$ such that $B_{r_1}(1) \subset [0,1]\cup [2,3] \setminus F$ and choose $r_2>0$ such that $B_{r_2}(2) \subset [0,1]\cup [2,3]\setminus F$. Let $r = \min\{r_1,r_2\}$. Now, $B_r(y) \subset [0,2]\setminus p(F)$. (To show this more rigorously, you should send $B_r(1) \cup B_r(2)$ through $p$ to show you get $B_r(y)$).

This shows $[0,2] \setminus p(F)$ is open, so $p(F)$ is closed.