in Set Theory & Algebra recategorized by
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29 votes
29 votes

Let $S$ be an infinite set and $S_1 \dots , S_n$ be sets such that $S_1 \cup S_2 \cup \dots \cup S_n = S$. Then

  1. at least one of the sets $S_i$ is a finite set
  2. not more than one of the sets $S_i$ can be finite
  3. at least one of the sets $S_i$ is an infinite
  4. not more than one of the sets $S_i$ can be infinite
  5. None of the above
in Set Theory & Algebra recategorized by
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4 Comments

@vaibhav101 No such assumption taken. @Niraj Singh 2  Yes n is finite.

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given Let S be an infinite set

S$i$ can be finite or infinite, but since we don't know what the other sets are, we'll treat S$i$ as infinite in order to make the final set S infinite as well.

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$\color{red}{\text{Finite Union of Finite Sets is Finite.}}$ i.e. If we have finite number of finite sets, then their union will necessarily be finite.

Union of $S_1,S_2, \dots, S_n$ is infinite set $\text{iff}$ at least one $S_1,S_2, \dots, S_n$ is infinite set. 

$\color{blue}{\text{Find detailed video solution here: }}$ Detailed Video Solution

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3 Answers

35 votes
35 votes
Best answer
  1. At least one of the set $S_{i}$ is a finite set. Well, it is not said that$S_{1},S_{2}\ldots S_{n}$ whether they are finite or infinite. It is possible to break down infinite sets into few sets (Some of which can be finite). This seems true, but I'm not able to prove it. Please Give a suitable counterexample here, if you think this is false.

          Ex-$: a^{\ast},$ this is infinite set. I can write it as $\{\}\cup \{a^{\ast}\},$ where $\{a^{\ast}\}$ is infinite.

  1. Not more than one of the sets can be finite. This is false.

         Ex $: a^{\ast}b^{\ast}\Rightarrow \{ab\} \cup  \{\} \cup \{{aa}^{+}{bb}^{+}\}.$

  1. At least one of the sets is Infinite. This must be True. As this is a finite union of sets, one of the sets must be infinite to make the whole thing infinite. True.
  2. Not more than one of the sets $S_i$ can be infinite. This is false. 

         Ex $: a^{\ast}b^{\ast} = \{a^{p}b^{q}|p=q\}\cup \{a^{m}b^{n}|m\neq n\}$ such that $p,q,m,n \geq 0.$

Answer C is surely true.

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4 Comments

Half of the elements will be even, and the other half will be odd, sir. Giving it one-to-one correspondence reduces it to a finite set.

This natural number example does not fit into statement A.
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It's not very difficult to prove option A as False because our task is simple, just to find a case where option A will be false. And indeed we have such a case: If all of $S_{1},S_{2}\ldots S_{n}$ are infinite sets and Infinite Union is Infinite, it becomes false as none of the sets are finite.
While in option C, its natural that in order for union to be infinite, at least one of the sets must be an infinite set. Therefore, Ans is Option C.

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For option a counter example case is take all finite set then S can never be infinite.
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11 votes
11 votes
atleast one of set should  b infinite..
9 votes
9 votes

If an infinite set is partitioned into finitely many subsets, then at least one of them must be infinite.

https://en.wikipedia.org/wiki/Infinite_set#Properties

Answer:

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