# Show that $X$ is countable.

Let $X \subset \mathbb{R}_{\geq 0 }$. Suppose there exists $C>0$ such that for any finite subset $\{x_1,x_2,...,x_n\}\subset X$ $\sum_{i=1}^{n}x_i \leq C$. Show that $X$ is countable.

I am quite lost in trying to solve this exercise. I've only thought about $\mathcal{P}_{<\infty}(X)=\{A\subset X:\#(A)<\infty\}$, and I found out that $\#(\mathcal{P}_{<\infty}(X))=\aleph_{0}$ if $\#(X)=\aleph_{0}$ and $(\mathcal{P}_{<\infty}(X))=\mathfrak{c}$ if $\#(X)=\mathfrak{c}$. However I can't go any further, and I don't even know if this is useful somehow. Any hint?

• Hint: Try to show that $X \cap [\frac{1}{k}, \infty)$ is finite for each $k \in \mathbb N$.
– user99914
May 18 '15 at 4:05
• I am more than certain that this question was asked before, more than once. I'm too busy to find a duplicate, at this moment, though. May 18 '15 at 5:31

Consider $$X_n=X\cap(1/n,\infty).$$ Then $\bigcup_{n=1}^\infty X_n=X$. If all the $X_n$'s were countable, then so would be $X$. Hence, for some $m\in\mathbb N$, we have that $\lvert X_{m}\rvert>\aleph_0$. Let $x_1,\ldots,x_N\in X_m$, where $N>cm$. Then $$x_1+\cdots+x_N>\frac{N}{m}>c.$$

• I don't see where you are using the uncountability of $X_m$ May 18 '15 at 4:34
• @Joaquin: Yiorgos is using the well-known fact that a countable union of countable sets is countable. Hence, by contraposition, if $X$ is uncountable, then $X_{n}$ is uncountable for some $n \in \Bbb{N}$. May 18 '15 at 5:30
• A more transparent way to argue might be: "If all the $X_n$s were finite, then $X$ would be countable (a countable union of finite sets); so some $X_m$ is infinite, so we can pick $>cm$-many elements of $X_m$, and the sum of these (finitely many) elements must be $>c$." It isn't important for $X_m$ to be uncountable, it just needs to be infinite. There's an interesting dichotomy here: in order for us to know that some $X_m$ is infinite, we need $X$ to be uncountable, at which point we know that some $X_m$ is actually not just infinite, but uncountable. May 18 '15 at 5:36
• Yes, that was the problem I was having when trying to write the solution (once they gave me the hint.) Because the $X_{n}$ could all be countable, and that would also work!. So the essential difference is that the sets $X_n$ are finite May 18 '15 at 13:08

Assume some point of $X$ is an interior point, $x$. Fix $r>0$ such the neighborhood of radius $r$ around $x$ is contained in $X$. This neighborhood has infinitely many points of $X$.

Choose $k$ such that $kx>C$, then pick $k-1$ many $x_i$ from the neighborhood around $x$ such that each $x_i>x$. Then the sum of all $k$ elements is greater than $C$, thus by contradiction, no point of $X$ can be an interior point.

Then any point in $X$ must be either an isolated point or a limit point of $X$ (or both). The set of isolated points is obviously countable, so the question becomes how many limit points there are.

For any nonzero limit point, there are infinitely many points in $X$ in any neighborhood of that limit point. Thus, by a similar process as above, we can create another contradiction to show that no such limit point exists. Of course, a limit point at $0$ is still possible, but that is not a problem in terms of countability.