Identifying the accumulation points of a set

real-analysis

Kindly bear with me as I am a beginner in this topic.

The goal is to identify the accumulation points of the set $\{5^a + 7^b: a, b \in \mathbb Z\}$. My definition of accumulation points is:

Definition: Let $X$ be a metric space with distance $d$. Then $x \in X$ is an accumulation point of $S \subseteq X$ if for all $\varepsilon> 0$ the ball $B(x,\varepsilon)$ contains at least one point in $S$ other than $x$.

In the case I am considering, $X = \mathbb{R}$ and the metric is the usual Euclidean distance (though I was not specifically told to use this metric).

With this definition in mind, it seems like the accumulation points of the set are just the points in the set itself. It seems definitely possible to choose $x\in \mathbb{R}$, $x\ne 5^a + 7^b$, and find $\varepsilon$ sufficiently small so that $B(x, \varepsilon)$ doesn't contain any number of the form $5^a + 7^b$—I'm thinking we could even pick $x\notin\mathbb Z$ and $\varepsilon$ very small so that $B(x, \varepsilon)$ wouldn't even contain an integer (and thus wouldn't contain any number of the form $5^a + 7^b$). Is this on the right track?

Would the isolation points just be everything in $\mathbb R$ that's not a number of the form $5^a + 7^b$?

Best Answer

A point $x\in\mathbb R$ is an accumulation point of

$$ X:=\{5^a+7^b \hskip1ex : \hskip1ex a,b \in \mathbb Z\}\, $$

if and only if there exists two sequences $(a_n), (b_n)$ in $\mathbb Z$ such that

$$ \{5^{a_n} + 7^{b_n}\;:\;n\in\mathbb N\}\,\subseteq X-\{x\} \qquad \text{and}\qquad 5^{a_n} + 7^{b_n} \;\underset{n}\longrightarrow\; x. \tag{1}\label{successione originale}$$

In this hypothesis it isn't possible that $(a_n)$ is upper unbounded. If were so, in fact, we would have that the left side of the limit in \eqref{successione originale} would be upper unbounded (being $7^{b_n} > 0 $), and therefore it could not converge to $x$. Similarly for $(b_n)$.

On the contrary, it can happen that $(a_n)$ or $(b_n)$ is lower unbounded. In the first case, we can extract from $(a_n)$ a subsequence $(a_{\sigma(n)})$ that diverges to $-\infty$, from which it follows $5^{a_{\sigma(n)}} \underset{n}\longrightarrow 0$. Now \eqref{successione originale} implies

$$ 5^{a_{\sigma(n)}} + 7^{b_{\sigma(n)}} \;\underset{n}\longrightarrow\; x, $$

and therefore $\;7^{b_{\sigma(n)}} \;\underset{n}\longrightarrow\; x$. Two cases can arise here:

  1. $(b_{\sigma(n)})$ is lower unbounded too. In this case (for brevity writing $\sigma\tau$ instead of $\sigma\,\circ\,\tau$) a subsequence $\;b_{\sigma\tau(n)}\;$ of $\;b_{\sigma(n)}$ diverge to $-\infty$, and therefore $7^{b_{\sigma\tau(n)}} \;\underset{n}\longrightarrow 0$, so $\;x=0$. And $x=0$ is actually an accumulation point of $X$ because for $n\in\mathbb N$ we have, for example, $\;0\neq 5^{-n}+7^{-n} \;\underset{n}\longrightarrow\; 0$.
  2. $(b_{\sigma(n)})$ is bounded. In this case the set of values ​​assumed by $b_{\sigma(n)}$ is finite being a bounded subset of $\mathbb Z$, and therefore there is a ${\mathit constant}$ subsequence $b_{\sigma\tau(n)}$ of $b_{\sigma(n)}$ which causes that $7^{b_{\sigma\tau(n)}}$ converge to $7^{\text{such a ${\mathit constant}$}}$. We deduce from what we have $x=7^b$, for some $b\in \mathbb Z$. And any $7^b, b\in \mathbb Z,$ is actually an accumulation point for $X$, because for $n\in\mathbb N$ we have, for example, $\;7^b \neq 5^{-n}+7^b\;\underset{n}\longrightarrow \;7^b$.

In a similar way, we conclude that, for any $a\in\mathbb Z,$ $5^a$ is an accumulation point for $X$.

The last case that remains to be examined is that which occurs when both $(a_n)$ and $(b_n)$ are bounded. In this case $(a_n)$ and $(b_n)$ only take a finite number of values, and therefore $5^{a_n} + 7^{b_n}$ also do the same. Consequently, $x$ must be exactly one of these values, which is assumed by the sequence for infinite values ​​of the index n; so we conclude that the sequence $5^{a_n}+7^{b_n}$ does not definitively belong to $X-\{x\}$, and therefore it cannot produce any accumulation point of $X$.

As there are no other possibilities, we conclude that the derived set $D(X)$ of $X$, i.e. the set of accumulation points of $X$, is given by

$$ D(X) = \{5^a \; : \: a\in\mathbb Z\}\,\cup\, \{7^b \; : \; b\in \mathbb Z\}\,\cup\,\{0\}. $$

Finally, the set $X^*$ of isolated points of $X$ is given by

$$ X^* = X-D(X) = X $$

since $X\,\cap\,D(X) = \emptyset.$

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