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Suppose $M$ is a metric space and $ K\subset N \subset M $, where $N$ is dense in $M$ and $K$ is dense in $N$, with both $K$ and $N$ being countable. Is it true that $K$ is dense in $M$?

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  • $\begingroup$ You're assuming that $K$ is dense in $M$. Is that assumption supposed to be "$K$ is dense in $N$"? $\endgroup$ – Chris Eagle Mar 11 '12 at 17:53
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The answer is yes and is true regardless of the cardinality of $K$ and $N$, and also regardless of whether $M$ is a metric space or not.

Take a nonempty open $U\subseteq M$. $N$ is dense in $M$ so $U$ contains a point $x\in N$. Now let $V\subseteq U$ be nonempty, open and contain $x$. Since $K$ is dense in $N$ there is a point of $K$ in $V$. This point is also in $U$. Therefore $K$ is dense in $M$.

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The countability doesn't really matter so much here, and neither does the fact we are talking about metric spaces.

Recall that a subset $D$ of a topological space $X$ is dense in $X$ iff $D \cap U \neq \emptyset$ for every nonempty open $U \subseteq X$. Also recall that if $Y$ is a subspace of a topological space $X$, then a subset $V \subseteq Y$ is open in $Y$ iff there is an open $U \subseteq X$ such that $V = U \cap Y$.

Letting $U$ be any nonempty open subset of $M$, by density of $N$ in $M$ we know that $U \cap N \neq \emptyset$. But it then follows that $U \cap N$ is a nonempty open subset of $N$. Since $K$ is dense in $N$ we have that $K \cap ( U \cap N ) \neq \emptyset$, which immediately implies that $K \cap U \neq \emptyset$.

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Assuming that you actually mean $K$ dense in $N$, the answer to your question is yes. Let $\epsilon > 0$ and $m\in M$ be arbitrary. We need to show there is a $k\in K$ with $d(k,m)<\epsilon$. Because $N$ is dense in $M$, there exists some $n\in N$ with $d(n,m)<\frac{\epsilon}{2}$. By the density of $K$ in $N$, there exists some $k\in K$ with $d(k,n)<\frac{\epsilon}{2}.$ Now use the triangle inequality: $d(k,m) \leq d(k,n) + d(n,m) < \epsilon$, and you're done.

Edit: Let me add that this is true in general topological spaces, not just in metric spaces. Patrick's answer is superior because he proves this in the setting of general topological spaces.

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Yes. Let $x\in M,\epsilon>0$. Then we have some $y\in N$ such that $d(x,y)<\epsilon/2$ because $N$ is dense in $M$, and some $z\in K$ such that $d(y,z)<\epsilon/2$ because $K$ is dense in $N$ (I assume you are using the subspace metric for $N$, which justifies not distinguishing between the two metrics as they agree wherever they are defined). By the triangle inequality, $$d(x,z)\leq d(x,y)+d(y,z)=\epsilon/2+\epsilon/2=\epsilon$$ hence $K$ is dense in $M$.

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Using the definition that the closure of a set $A$ is the intersection of all closed sets containing $A$: The closure of $K$ in $M$ is a closed set containing $N$, and thus contains the closure of $N$ in $M$.

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