# Munkres Topology Ch. 1 section 2 Ex #1

• benorin
Also, note that in general, if you have a function ##g:A\to B## and a set ##C\subset A##, then ##g(C)=\{g(c):c\in C\}## is well-defined and unambiguous notation. But if you have a set ##D\subset B## and want to talk about ##g^{-1}(D),## this might be a little unclear. Do you mean the set of all elements in ##A## that map to elements of ##D##? Or do you mean the set of all elements in ##A## that map to elements of ##D## and are also in the range of ##g?## For example, if ##g:\mathbb{R
benorin
Homework Helper
Homework Statement
If ##f: A\rightarrow B## and ##A_0\subset A## and ##B_0\subset B##. (a) show that ##A_0\subset f^{-1}(f(A_0))## and that equality holds if ##f## is injective. (b) show that ##f(f^{-1}(B_0))\subset B_0## and that equality holds if ##f## is surjective.
Relevant Equations
The definitions of injective and surjective. Also definitions of ##f## and ##f^{-1}## restricted to a subset of the domain or range. I think I understand this part on my own but will type these up below if you want me to, I’m on my phone and TeX is a pain.
Mostly I need to clear up a few basic things about functions and their inverses, the problem seems easy enough. Ok, so for (a) I have

$$f^{-1}(f(A_0))= \left\{ f^{-1}(f(a)) | a\in A_0\right\}$$

but here I’m not certain if ##f^{-1}## is allowed to be multi-valued or not, the text says that if ##f## is bijective then ##f^{-1}## exists and is also bijective. But it didn’t specifically say “if, and only if” just if. And clearly the problem stipulates the existence of the inverse of ##f## even if it is not even injective so I’m unclear as to the nature of the inverse function here? I hope that makes sense.

Last edited:
If ##f## is a function from ##A## to ##B## and if ##C\subset B##, then ##f^{-1}(C)## means, just by definition, the set ##\{a\in A: f(a)\in C\}.## It is not the image of ##C## under a function called ##f^{-1}##, and we are not assuming that ##f## is invertible.

With this in mind,
benorin said:
$$f^{-1}(f(A_0))= \left\{ f^{-1}(f(a)) | a\in A_0\right\}$$

doesn't make sense. In your RHS, you are applying ##f^{-1}## to an element of ##B## rather than a subset of ##B##. Even if you were to write ##\{f^{-1}(\{f(a)\}):a\in A_0\}## on your RHS (and it is indeed a common 'abuse of notation' to write ##f^{-1}(x)## in place of ##f^{-1}(\{x\})##), it would still be off. For example, suppose ##f:\{0,1\}\to \{0,1\}## is just the constant function ##f(0)=f(1)=0.## Then your RHS is ##\{f^{-1}(0)\}=\{\{0,1\}\},## which is not a subset of the domain, whereas your LHS is ##f^{-1}(\{0\})=\{a\in \{0,1\}: f(a)\in\{0\}\}=\{a\in \{0,1\}: f(a)=0\}=\{0,1\}.##

A correct expression along these lines would be ##f^{-1}(f(A_0))=\bigcup_{a\in A_0} f^{-1}(\{f(a)\}),## but this doesn't really get you closer to a solution. Instead, try to show that ##A_0## is always a subset of ##f^{-1}(f(A_0))## by taking an element of ##A_0## and showing that it must be in ##f^{-1}(f(A_0)).##

Infrared said:
If ##f## is a function from ##A## to ##B## and if ##C\subset B##, then ##f^{-1}(C)## means, just by definition, the set ##\{a\in A: f(a)\in C\}.## It is not the image of ##C## under a function called ##f^{-1}##, and we are not assuming that ##f## is invertible.

This^ helps, thanks. Now that I'm on my Mac (and not my phone) I will write the step I skipped earlier, namely

$$\boxed{f^{-1}(f(A_0))= \left\{ f^{-1}(b) | b\in f(A_0)\right\} } =\left\{ f^{-1}(f(a)) | a\in A_0\right\}$$
I now know that the second equality doesn't make sense given what you said. I did think that ##\{ b | b\in f(A_0)\} = \{ f(a) | a\in A_0\}## was an ok simplification. I guess not.

Infrared said:
With this in mind,doesn't make sense. In your RHS, you are applying ##f^{-1}## to an element of ##B## rather than a subset of ##B##. Even if you were to write ##\{f^{-1}(\{f(a)\}):a\in A_0\}## on your RHS (and it is indeed a common 'abuse of notation' to write ##f^{-1}(x)## in place of ##f^{-1}(\{x\})##), it would still be off. For example, suppose ##f:\{0,1\}\to \{0,1\}## is just the constant function ##f(0)=f(1)=0.## Then your RHS is ##\{f^{-1}(0)\}=\{\{0,1\}\},## which is not a subset of the domain, whereas your LHS is ##f^{-1}(\{0\})=\{a\in \{0,1\}: f(a)\in\{0\}\}=\{a\in \{0,1\}: f(a)=0\}=\{0,1\}.##

A correct expression along these lines would be ##f^{-1}(f(A_0))=\bigcup_{a\in A_0} f^{-1}(\{f(a)\}),## but this doesn't really get you closer to a solution. Instead, try to show that ##A_0## is always a subset of ##f^{-1}(f(A_0))## by taking an element of ##A_0## and showing that it must be in ##f^{-1}(f(A_0)).##

From the usage in the text, I've inferred that ##f^{-1}## may be applied to both an element of ##B## and a subset of ##B##, where context is key, as I have seen both explicitly; is the former only valid if ##f## is invertible?

benorin said:
I did think that ##\{ b | b\in f(A_0)\} = \{ f(a) | a\in A_0\}## was an ok simplification. I guess not.
This is basically correct, but not what you did in your OP. The LHS is a little clunky though- why would you write ##\{x: x\in E\}## instead of just ##E.##?

In general though, when writing sets in this way, it's better to write something like ##\{x\in C: \Phi(x)\}## instead of ##\{x:\Phi(x)\}## where ##\Phi(x)## is some predicate. If you don't include a domain, then you can get issues like Russel's paradox (thinking about the "set" ##\{x:x\notin x\}## leads to contradictions) as well as just annoying ambiguities, e.g. in ##\{x: x^3=1\}##, is ##x## supposed to be a real number, or a complex number, or maybe a matrix...? Here it's clear what you mean, but just be careful.
benorin said:
From the usage in the text, I've inferred that ##f^{-1}## may be applied to both an element of ##B## and a subset of ##B##, where context is key, as I have seen both explicitly; is the former only valid if ##f## is invertible?
If ##f## is invertible, then ##f^{-1}(b)## makes sense- it is just the function ##f^{-1}## applied to the element ##b.## If ##f## is not necessarily invertible, then to be perfectly strict, you should only write ##f^{-1}(C)## where ##C## is a subset of ##B##, but sometimes ##f^{-1}(b)## is written as shorthand for ##f^{-1}(\{b\})##, and this latter expression is well-defined (it's the set of all elements of ##A## that map to ##b##).
For example, if ##f:\mathbb{R}\to\mathbb{R}## is the function ##f(x)=x^2##, then ##f^{-1}(1)=f^{-1}(\{1\})=\{1,-1\}## according to this notation. ##f^{-1}## is not a function on its own. It might be a good idea to avoid this shorthand if you're still getting used to this stuff.

member 587159

## 1. What is the purpose of Munkres Topology Ch. 1 section 2 Ex #1?

The purpose of this exercise is to introduce the concept of topological spaces and their basic properties, such as open and closed sets, interior and closure, and limit points.

## 2. What are some key definitions and concepts covered in this exercise?

This exercise covers definitions such as topological spaces, open and closed sets, interior and closure, and limit points. It also introduces the concept of a basis for a topology and the definition of a subspace topology.

## 3. How does this exercise relate to real-world applications?

Topology is a fundamental branch of mathematics that has many applications in various fields, such as physics, engineering, and computer science. Understanding the basic concepts covered in this exercise is essential for further study and application of topology in these fields.

## 4. What are some common mistakes made by students when solving this exercise?

Some common mistakes made by students include confusing open and closed sets, not understanding the definition of a basis for a topology, and not being able to correctly identify limit points.

## 5. How can this exercise be further extended or applied?

This exercise can be further extended by exploring more complex topological spaces and their properties, such as compactness, connectedness, and separation axioms. It can also be applied in various fields, such as data analysis, network theory, and geometry.

• Calculus and Beyond Homework Help
Replies
1
Views
493
• Math Proof Training and Practice
Replies
1
Views
946
• Calculus and Beyond Homework Help
Replies
6
Views
354
• Calculus and Beyond Homework Help
Replies
3
Views
538
• Calculus and Beyond Homework Help
Replies
11
Views
1K
• Calculus and Beyond Homework Help
Replies
15
Views
2K
• Calculus and Beyond Homework Help
Replies
15
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Precalculus Mathematics Homework Help
Replies
4
Views
2K