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I know that for this problem I have to use contradiction. Could anyone check my work and guide me through the problem if it's wrong? So far, this is what I have.Thanks!!

Contradiction: $\mathcal B\subseteq \mathcal A$, then $\mathcal B$ is not a family of pairwise disjoint sets.

If $\mathcal B$ is not pairwise disjoint then $\mathcal A \neq \mathcal B$ or $\mathcal A \cap \mathcal B\neq \varnothing $

then, x $\in \mathcal A$ and x$\notin \mathcal B$

However, x $\in \mathcal B\subseteq \mathcal A$ , which is a contradiction since we said that x$\notin \mathcal B$.

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    $\begingroup$ This does not prove the claim. Two things you should look into: 1. $\mathcal{A}$ (and $\mathcal{B}$) is a set of sets and no pair of these sets have common members; 2. In the argument $x$ is a free variable at some point, but not in an other. $\endgroup$ – AD. Jun 24 '13 at 5:13
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There is no need to use contradiction. Suppose that $B_0,B_1$ in $\mathscr{B}$ with $B_0\ne B_1$. Then $B_0$ and $B_1$ are distinct members of $\mathscr{A}$, so $B_0\cap B_1=\varnothing$ (since $\mathscr{A}$ is a pairwise disjoint family). Thus, every pair of distinct members of $\mathscr{B}$ are disjoint, and by definition $\mathscr{B}$ is a pairwise disjoint family.

The argument that you’ve given does not make sense. $\mathscr{B}$ is a family of sets, and in order to show that this family is pairwise disjoint, you must show that if $B_0,B_1\in\mathscr{B}$, then either $B_0=B_1$, or $B_0\cap B_1=\varnothing$. There is no point to looking at $\mathscr{A}\cap\mathscr{B}$: we know that $\mathscr{A}\cap\mathscr{B}=\mathscr{B}$ simply because $\mathscr{B}\subseteq\mathscr{A}$. This fact has in itself no bearing on whether $\mathscr{B}$ is pairwise disjoint.

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  • $\begingroup$ Would you mind to do approach the problem as a contradiction? $\endgroup$ – Kururugi Suzaku Jun 24 '13 at 5:01
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    $\begingroup$ @KururugiSuzaku: It’s an unnecessary complication, but it’s easy enough to do. If $\mathscr{B}$ is not a pairwise disjoint family, then there are $B_0,B_1\in\mathscr{B}$ such that $B_0\ne B_1$ and $B_0\cap B_1\ne\varnothing$. But $\mathscr{B}\subseteq\mathscr{A}$, so $B_0,B_1\in\mathscr{A}$, contradicting the pairwise disjointness of $\mathscr{A}$. $\endgroup$ – Brian M. Scott Jun 24 '13 at 5:03
  • $\begingroup$ Alright, after looking at the definition it makes sense. Thanks :) $\endgroup$ – Kururugi Suzaku Jun 24 '13 at 5:23
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    $\begingroup$ @KururugiSuzaku: You’re welcome. (Until you’re thoroughly comfortable with some new concept, it’s probably a good idea to go back and check the definition whenever you have to prove something about the concept. Many elementary results pop right out of the definitions involved.) $\endgroup$ – Brian M. Scott Jun 24 '13 at 5:25

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