# Subsets of Uncountable Infinite Sets

• hetaeros
In summary: Thanks for the help!In summary, A - B is an infinite, uncountable set which has the same cardinality as A. This is shown by proving that (A-B) union B is countable, and then using this fact to construct a bijection between A and A-B. The proof uses the hint of defining a countable subset C of A-B and constructing a map from C union B to B. To prove that A-B is uncountable, a proof by contradiction is used, showing that if A-B were countable, then A would also be countable, leading to a contradiction with the given information that A is uncountable.
hetaeros

## Homework Statement

Let $$A$$ be an infinite set which is not countable and let $$B \subset A$$ be a countably infinite set.

(1) Show that $$A - B$$ is also infinite and not countable

(2) Show that $$A$$ and $$A - B$$ have the same cardinality

## Homework Equations

Hints written directly:

"Show that there is a countable subset $$C \subset (A - B)$$ and that there is a bijection $$g : C \cup B \rightarrow B$$, and try to make it into a bijection of $$A$$ with $$A - B$$."

## The Attempt at a Solution

Here is what I have so far:

Given the hint, I am thinking that I want to use the fact that a set is infinite if there is a bijection between it and a strict subset of it. However, there's a possibility that the set $$C$$ is a finite subset, which is why I believe the professor gave us that hint about constructing a map from $$C \cup B$$ to make sure it is infinite.

Here, then is where I show that $$C \cup B$$ is infinitely countable.

Define $$C \subset (A - B)$$ as a countable subset of $$A - B$$
Construct $$f : C \cup B \rightarrow B$$
Since $$C$$ and $$B$$ are disjoint countable sets, their union is also countable.

$$B$$ is a countably infinite set
$$\Rightarrow C \cup B$$ is countably infinite
$$\Rightarrow #B = #(C \cup B) = \aleph_{0}$$
$$\exists bijection f : C \cup B \rightarrow B$$

However, this is where I get stuck. I can easily conclude that since $$A$$ is not countable, there is no bijection between and $$C \cup B$$, but that seems quite trivial at the moment, especially if I'm going to be using the theorem from above.

The part that gets me the most is how $$C \cup B$$ could possibly be a strict subset of $$A - B$$, since $$B \notin (A - B)$$. I will probably need to use this fact that $$A - B$$ is infinite, but I am completely lost as to how to go about this.

To prove that $$A - B$$ is also uncountable: will I need to show that there cannot exist a bijection between it and $$C \cup B$$ ? The funny thing about this is that if $$A - B$$ is infinite, it must have a bijection between it and $$C$$, which is where the whole issue with $$C \cup B$$ turns almost circular.

Can someone please help me clear things up and perhaps find a better strategy for a proof?

As per part (2), I would assume that the easiest way to do this is to show that there exists a bijection between $$A - B$$ and $$A$$. Would the results of the previous part be required to solve this?

Thank you for your time--any help would be greatly appreciated!

For part (1), just notice $$(A-B) \cup B$$ = A, and the union of two countable sets is countable.

For part (2), you (should) know there is a countable subset C of A-B. This is useful to know. When making your bijection from A to A-B, try separating it into the case of whether an element of A is in (A-B)-C or not.

Office_Shredder said:
For part (1), just notice $$(A-B) \cup B$$ = A, and the union of two countable sets is countable.

For part (2), you (should) know there is a countable subset C of A-B. This is useful to know. When making your bijection from A to A-B, try separating it into the case of whether an element of A is in (A-B)-C or not.

Ah, but we're trying to prove that $$(A - B)$$ is uncountable.

Office_Shredder is gently suggesting a proof by contradiction. If (A-B) were countable then so would (A-B) union B. But that's A. And it's not countable. Hence A-B is uncountable.

Ah, I see. The contradiction comes from the fact that the given A is uncountable.

## 1. What are subsets of uncountable infinite sets?

Subsets of uncountable infinite sets are a group of elements that are contained within a larger set that is infinite and cannot be counted. These subsets can be either countable or uncountable themselves, and they may have different properties and characteristics compared to the larger set.

## 2. How do subsets of uncountable infinite sets differ from subsets of countable infinite sets?

Subsets of uncountable infinite sets differ from subsets of countable infinite sets in that they have different cardinalities. While subsets of countable infinite sets have the same cardinality as the larger set, subsets of uncountable infinite sets have a smaller cardinality. This means that there are fewer elements in the subset compared to the larger set.

## 3. Can subsets of uncountable infinite sets be countable?

Yes, subsets of uncountable infinite sets can be countable. This means that the elements in the subset can be put into a one-to-one correspondence with the natural numbers. However, the subset will still have a smaller cardinality compared to the larger uncountable infinite set.

## 4. What are some examples of subsets of uncountable infinite sets?

Some examples of subsets of uncountable infinite sets include the set of real numbers, the set of irrational numbers, and the set of all functions from the natural numbers to itself. These subsets have different properties and characteristics compared to the larger sets, and they can be studied and analyzed separately.

## 5. How are subsets of uncountable infinite sets used in mathematics?

Subsets of uncountable infinite sets are used in mathematics to study the properties and characteristics of these sets in a more manageable way. By examining subsets, mathematicians can better understand the behavior and structure of these larger sets. Subsets of uncountable infinite sets also have many applications in other areas of mathematics, such as in topology and analysis.

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