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Let $(\Omega, d)$ be a metric space, and let $A,B \subseteq \Omega$ such that

  1. $A \subseteq \overline{B}$ and
  2. $A$ is dense in $\Omega$.

Show that $B$ is also dense in $\Omega$.

Here is my attempt at a solution:

Let $x \in \overline{B}$. Then there is a sequence $(x_n)_{n \in \mathbb{N}} \in B$ such that $x_n \to x$. Since $A \subseteq \overline{B}$, then $x_n \to a$ for some $a \in A$. We are also given that $A$ is dense in $\Omega$, so $d(x,A) = 0$. But $x \in \overline{B}$, so $d(x, \overline{B}) = 0$ too.

In the following sentence, I quote a Lemma from my textbook that reads: Let $(\Omega, d)$ be a metric space, and let $A , B \subseteq \Omega$. Then the following assertions hold:

a). $\overline{\emptyset} = \emptyset, \quad \overline{\Omega} = \Omega$.
b). $ A \subseteq \overline{A}$.
c). $A \subseteq B \Rightarrow \overline{A} \subseteq \overline{B}$.

I quote part (b) by stating this:

$B \subseteq \overline{B}$, so $d(x, \overline{B}) = 0 \Rightarrow d(x, B)=0$ which means that $B$ is dense in $\Omega$.

As an alternative solution, instead of going off of part (b) of this Lemma, I consider using part (c) as follows:

If $\overline{A} \subseteq \overline{B}$, then $\Omega \subseteq \overline{B}$ since we are given that $A$ is dense in $\Omega$. So $\Omega \subseteq \overline{B}$ and $B \subseteq \Omega$, so we conclude that $B$ is dense in $\Omega$ since its closure is equal to the entire space.

Is this the right way to approach this problem? Any suggestions or hints will be greatly appreciated. Much thanks in advance for your time and patience.

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    $\begingroup$ The easiest solution is to do as you've done and quote (c), since if $A$ is dense in $\Omega$, then by (c) $\Omega = \overline A \subset \overline B \subset \Omega$, so $\overline B = \Omega$ $\endgroup$ – Mathmo123 Mar 1 '15 at 0:14
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Your alternative solution is correct and is definitely the way to go. There are some problems with your first attempt; I’ll quote part of it with comments.

Let $x \in \overline{B}$.

You’re trying to show that $B$ is dense in $\Omega$, so this is the wrong place to start: if you’re going to use this sequences approach, you need to start with an arbitrary $x\in X$ and show that there’s a sequence in $B$ that converges to it.

Then there is a sequence $(x_n)_{n \in \mathbb{N}} \in B$ such that $x_n \to x$.

True, but (as noted above) not really useful.

Since $A \subseteq \overline{B}$, then $x_n \to a$ for some $a \in A$.

This is not true unless $x$ happens to be in $A$. Since $A$ may well be a proper subset of $\overline{B}$, there’s no reason to suppose that $x\in A$.

It’s possible to make this kind of argument work, however. Let $x\in X$ be arbitrary. $A$ is dense in $X$, so there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $A$ that converges to $x$. $A\subseteq\overline{B}$, so $x_n\in\overline{B}$ for each $n\in\Bbb N$. Thus, for each $n\in\Bbb N$ there is a sequence $\langle x_{n,k}:k\in\Bbb N\rangle$ in $B$ that converges to $x_n$. To complete the proof that $B$ is dense in $\Omega$, you need to extract from that collection of sequences one that converges to $x$. This is possible but a bit messy.

For each $n\in\Bbb N$ there are $k(n),\ell(n)\in\Bbb N$ such that $$d(x_{k(n)},x)<\dfrac1{2^{n+1}}$$ and $$d(x_{k(n)},x_{k(n),\ell(n)})<\dfrac1{2^{n+1}}\;,$$ so that $$d(x_{k(n),\ell(n)},x)<\frac1{2^n}$$

by the triangle inequality. Clearly the sequence $\langle x_{k(n),\ell(n)}:n\in\Bbb N\rangle$ converges to $x$ and lies entirely in $B$.

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  • $\begingroup$ Thank you very much for the elaborate comments! I appreciate your explanations as to why my first attempt was not correct as it stands. I will definitely go with the alternate approach. $\endgroup$ – Jamil_V Mar 1 '15 at 5:23
  • $\begingroup$ @Jamil: You're very welcome. $\endgroup$ – Brian M. Scott Mar 1 '15 at 6:51

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