Virginia Tech 2019 question Last week I took the Virginia Tech math contest, and the following was a problem: 
Let S be a subset of $\mathbb{R}$ with the property that for every $s\in S$, there exists $\epsilon > 0$ such that $(s-\epsilon, s + \epsilon) \cap S = \{s\}$. Prove that there exists a function $f: S \rightarrow \mathbb{N}$ that is one-to-one. 
I eventually figured out that one must simply use the density of rationals and define a function from S to $\mathbb{Q}$ and take the composition of that function with a function from $\mathbb{Q}$ to $\mathbb{N}$. That seems to be the standard answer. 
My question is whether you can prove this by way of contradiction. Can you suppose there is some function $\varphi: \mathbb{R} \rightarrow S$ that is 1-1 and get a contradiction? I know this is certainly not a preferred technique in set theory for cardinality. Is it possible to arrive at a contradiction? I certainly could not think of an obvious, explicit conclusion. 
 A: Let $\phi$ be such function, first I claim that there exists $\{a_s\}_{s\in S}$, a set of disjoint intervals with $a_s\cap s=\{s\}$.
Let $(s_i)_{i\in|S|}$ be well ordering of $S$, and then let $a_{s_i}$ be $(s_i-\epsilon, s_i+\epsilon)$ such that it is disjoint with $a_{s_j}$ for $j<i$, and $a_{s_i}\cap s_i=\{s_i\}$.(if it doesn't exist, shorten all the previous $a_{s_j}$ by $1/2$, repeat this process till you can define $a_{s_i}$, for you never can, you have a sequence in $S\setminus \{s_i\}$ that converge to $s_i$, so for each $\delta$ we have $(s_i-\delta, s_i+\delta)\cap S$ infinite, which is contradiction.).
But then, by letting $g:\Bbb R\to\{a_s\}_{s\in S}$ by $g(x)=a_{\phi(x)}$ you get uncountable many disjoint intervals.
By the fact that each $a_s$ contains at least 1 rational, you can define $h(x)=\mbox{a rational that is element of }g(x)$ you get that the rationals are uncountable.
Note that this only proves that $|S|<|\Bbb R|$, not that $|S|\le|\Bbb N|$, but replacing $\phi$ with any function $\psi:A\to S$ with $A$ is uncountable subset of $\Bbb R$ and $\psi$ is one to one will work, so just take $A$ to be a set with $|A|=\aleph_1$, and you get that $|S|\le|\Bbb N|$
A: Showing there is no injection or bijection from $\Bbb R$ to $S$ does not suffice because the axioms of set theory cannot prove nor disprove that there is an uncountable cardinal $K$ with $K<|\Bbb R|.$
My Answer to this Q Prove that an uncountable set has at least one accumulation point uses the density of $\Bbb Q$ in $\Bbb R$ to show that if $S$ (called "$A$" in that Q) is an uncountable subset of $\Bbb R$ then all but countably many members of $S$ are accumulation points of $S.$ 
The existence of an $x\in S$ such that $x$ is an accumulation point of $S$ would contradict the existence of an $\epsilon>0$ such that $(-\epsilon+x,\epsilon+x)\cap S=\{x\}.$
A: $\Bbb R$ has a countable base and so any subspace of $\Bbb R$ also has a countable base. From the assumption all $s \in S$ are isolated points and so any base for the topology of $S$ must contain all $\{s\}$. The combination gives that $S$ is at most countable: 
Any base has size $\ge|S|$  and there is a base of size $\le |\Bbb N|$ so $|S| \le |\Bbb N|$. Hence $f$ exists.
