Suppose that $X$ is a non-empty, complete metric space and $C_{n},n\in \mathbb{N}$, is a family of closed sets in $X$ such that $X=\cup_{n\in \mathbb{N}}C_{n}$. Prove that there exists $N\in \mathbb{N}$ such that $C_{N}$ has non empty interior.

Here is my attempt below.

My plan was to consider prove by contradiction and supposing that for all $n\in \mathbb{N}$ $C_{n}$ has empty interior. I then thought of considering an $x_1$ in some $C_{k_1}$ and considering an open ball with radius 1 and choosing a point $x_{2}$ in $C_{k_{2}}$. This must exist as I have postulated that the $C_{i}$ have empty interior for all $i\in\mathbb{N}$. I will continue this process indefinitely taking $x_{3}\in B(x_{2},r_{2})$ where $r_{2}$ is such that $B(x_{2},r_{2})\subset B(x_{1},1)$, taking $x_{4}\in B(x_{3},r_{3})$ where $r_{3}$ is such that $B(x_{3},r_{3})\subset B(x_{2},r_{2})$. In this way I have generated a Cauchy sequence $x_{n}$ which since the space is complete will converge to some $x$. From here I am kind of lost...Is this even the right way to attack the problem.


2 Answers 2


You're on the right track - the problem, though, is that you need the sequence you build to be "problematic": there has to be a reason it doesn't converge. Let me break my answer down into a series of hints/exercises:

Exercise 1: let $C$ be any set, and suppose I have a sequence $S=(a_i)$, and there is some closed ball $B$ such that all but finitely many elements of $S$ lie in $B$ and $B\cap C=\emptyset$. Then $S$ does not converge to an element of $C$.

So what does this have to do with the problem? Well, we want to build a sequence that is forced to stay away from each $C_n$ in turn. You've started on the right idea - if your first point was from $C_{k_1}$, you want your second point to be from some different $C_{k_2}$. But it requires more work. What you want is a Cauchy sequence $S=(a_i)$ which is guaranteed not to converge to any element of $C_n$, for any $n$. This will then give you a contradiction, since $\bigcup C_n=X$ and $X$ is complete.

Exercise 2: Using the previous exercise, show that it would be enough if you could find a sequence of points $S=(a_i)$ and a sequence of closed balls $T=(B_i)$ such that

  • the $a_i$s live in the $B_i$s: $i\ge j\implies a_i\in B_j$; and

  • the $B_i$s avoid the $C_i$s: $B_i\cap C_i=\emptyset$ for every $i$; and

  • the $B_i$s are small: the diameter of $B_i$ is at most $2^{-i}$.

To get started, show that if you find such an $S$ and $T$, then the sequence $S$ is Cauchy.

This, of course, leaves:

Exercise 3: Do that thing in Exercise 2. (HINT: this is where you will need to use the fact that each $C_i$ is closed. You'll prove a more general fact: that if $x\in C$ and $C$ is closed with no interior, then for any $\epsilon>0$ there is some point $y$ within $\epsilon$ of $x$ and some $\delta>0$ such that the closed ball of radius $\delta$ centered at $y$ is disjoint from $C$.)

  • $\begingroup$ Thankyou for your help. Here is my attempt of completing the original question with the use of your hints. Consider $C_{1}$ and suppose $\endgroup$
    – Damo
    Oct 26, 2015 at 13:03

Consider $C_{1}$ and let $y_{1}$ be a point not in $C_{1}$ then we know there will exist a $B_{1}(y_{1},\delta_{1})$ such that $B_{1}(y_{1},\delta_{1})$ will be disjoint from $C_{1}$, also choose $\delta_{1}<2^{-1}$. Now we consider $C_{2}$, since $C_{2}$ has no interior there will exist a point $y_{2}\in B_{1}(y_{1},\delta_{1})$ and some $\delta_{2}$ such that $B_{2}(y_{2},\delta_{2})\subset B_{1}(y_{1},\delta_{1})$ and does not intersect $C_{2}$, naturally $B_{2}$ will have empty intersection with $C_{1}$ as well, also we choose $\delta_{2}<2^{-2}$. Continuing in this process we have created a Cauchy sequence $(y_{i})$ which will converge to some $y\in X$ as $X$ is complete. However $y$ was constructed in such a way that it will not be contained in any $C_{i}$ which is a contradiction to the fact that $\cup C_{n}=X$.


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