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Let $S$ be a nonempty subset of $\mathbb{R}$ so that $S\ne\mathbb{R}$. There are two cases to consider: 1) $S^c$ is dense in $S$ ($\forall{s_1,s_2}\in{S}$ where $s_1<{s_2}$, ${\exists}s'\in{S^c}$ so that $s_1{<}s'{<}s_2$) and 2) $S^c$ is not dense in $S$.
 
If $S^c$ is dense in $S$, choose a member of $S$ and call it $s$. One can construct a sequence of members of $S^c$ that converges to $s$. Take a member of $S^c$ and call it $s_0$. For each $s_n$, if $\frac{s+s_n}{2}\in{S_c}$, make that number $s_{n+1}$. Otherwise, $\frac{s+s_n}{2}\in{S}$, and so choose a member of $S_c$ that is between $\frac{s+s_n}{2}$ and $s$, which exists because of the density of $S_n$ in $S$. This sequence converges to $s$, and because of that, s is a limit point of $S_c$. However, $s\notin{S_c}$, and since $S_c$ does not contain one of its limit points, $S_c$ is not closed, meaning that $S$ is not open.
 
If $S_c$ is not dense in $S$, then there must exist some interval in which everything is in $S$. Call this interval $T$. $T$ must be bounded from above or below because otherwise, $T$ would encompass all of $\mathbb{R}$. Therefore, either $\sup(T)$ or $\inf(T)$ exists.
Suppose $\inf(T)$ exists. Define $W=\{w\in{S_c}:w{<}\inf(T)\}$, and then define $T'=T$ $\cup$ $\{x\in{\mathbb{R}}:\sup(W){<}x{\leq}\inf(T)\}$ $\cup$ $\{x\in{S}:x=\sup(W)\}$. Because of the way in which $T'$ was constructed, $T'$ will only consist of members of $S$, and $\inf(T')=\sup(W)$. If $\inf(T')\in{T'}$, then $\inf(T')\in{S}$, and this implies that $S$ is not open. The reason for this is that for any $\epsilon$, there will be some $w\in{S_c}$ so that $\inf(T')-\epsilon{<}w{<}\inf(T')$ since $\inf(T')=\sup(W)$.

Thu, 18 Mar 2021 20:57 GMT