Simple topology problem involving continuity

[Math Amateur]

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

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 $X$ and $Y$ be topological spaces. A function $f \ : \ X \to Y$ is said to be continuous if for each open subset V of Y, the set $f^{-1} (V)$ 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 $p$ "is said to be a closed map if for each closed set $A$ of $X$ the set $p(A)$ is closed in $Y$"

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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

 

[HallsofIvy]

Okay, suppose U is open in [0, 2]. We want to show that $f^{-1}(U)$ is an open set in $[0, 1]\cup [2, 3]$.

Define $A= f^{-1}(U)\cap [0, 1]$ and $B= f^{-1}\cap [2, 3]$. If both A and B are open then their union, $f^{-1}(U)$ is open.

So we need to show that A and B are open. Let y be in U. Then either y= 1 or y= 0, or y= 2, or y is in (0, 1), or y is in (1, 2].

1) if y= 1, y= 1= f(1)= f(2)

2) if y= 0, y= 0= f(0)

3) if y= 2, y= 2= f(3)

Those are "endpoints" of intervals so any open set about them is of the form [0, x) or [x, 1) in [0, 1] or [x, 2) in [1, 2].

2) if y in (0, 1) then there exist x in (0, 1) such that f(x)= x= y.

3) if y in (1, 3) then there exist x in (2, 3) such that f(x)= x- 1= y.

 

[micromass]

In this case, you can check continuity like you did in analysis: with epsilon-delta definitions.

So in topology, if you encounter a map between metric spaces, you can just check the epsilon-delta definition of continuity. You don't need to work with open sets here. In fact, I think it's better to always avoid the open set definition whenever it is possible.

So if you know continuity from analysis well, then you should have no problems with the topological definition.

Closedness of maps is a different issue. There you indeed need to work with the closed sets. Luckily, there are some nice theorems which allow us to conclude that a map is closed. In fact, closedness as a concept is important because we care about these theorems. It's not like it's of independent interest, like continuity.

 

[Math Amateur]

Thanks HallsofIvy and micromass ... your posts were extremely helpful

Peter

 

[blue_raver22]

I can understand your confusion with the abstract definitions and how they can be applied in practice to demonstrate continuity and closedness of a map. Let's break down the definitions and see how they can be used in the example given by Munkres.

First, let's define what it means for a map to be continuous. In simple terms, a map is continuous if small changes in the input result in small changes in the output. In the context of topology, this means that the preimage of an open set in the codomain (Y) should be an open set in the domain (X). This is what the definition given by Munkres is saying.

Now, to prove the continuity of p in the example, we need to show that for any open set V in Y, the set p^{-1}(V) is an open set in X. Let's consider an open set V in Y, then by definition, we have p^{-1}(V) = {x in X | p(x) in V}. Now, since p is surjective, for every y in V, there exists an x in X such that p(x) = y. This means that p^{-1}(V) will contain all the points in X that map to points in V. Since V is an open set in Y, we can say that p^{-1}(V) is also an open set in X. This proves the continuity of p.

Similarly, to prove the closedness of p, we need to show that for any closed set A in X, the set p(A) is closed in Y. Let's consider a closed set A in X, then by definition, we have p(A) = {p(x) in Y | x in A}. Now, since p is surjective, for every y in Y, there exists an x in X such that p(x) = y. This means that p(A) will contain all the points in Y that are mapped from points in A. Since A is a closed set, all the points in X that map to points in A are also contained in A. This means that p(A) is a subset of A and hence a closed set in Y. This proves the closedness of p.

In summary, to prove the continuity and closedness of a map, we need to use the definitions and the properties of the map (in this case, surjectivity) to show that