Prove the cluster points for $A=(0,1)$ in $\mathbb{R}$ is the set $[0,1]$

Question

Let $\epsilon > 0$ be given and let $a \in [0,1]$ be a cluster point then we need to show $\exists x \in A$ such that $0 < |x-a| < \epsilon$

$|x-a| < \epsilon$ $\implies$ $a-\epsilon<x<a+\epsilon$

We also want $x \in A \implies$ $0 < x < 1$

Putting this together, we want

$x \in (max\{a - \epsilon, 0\}, min\{a + \epsilon, 1\})$

So by density of $\mathbb{R}$, there does indeed exist $x$ in the above interval. Also by the above condition, $x \in A$. So $[0,1]$ are the cluster points of $A$ in $\mathbb{R}$.

$\Box$

Is my proof correct? What can I improve in it?

I also thought about using the Archimedes property of reals to find an element in the above given interval. But that would mean elements of the form $1/n$, which are fine since they will still fit the definition of cluster points, but I thought I could be more generic and cover all types of elements by using density of reals.

Answer 1

There is no logic in your arguments.

Let $C$ be the set of all cluster points of the set $A=(0,1)$. What you are asked to prove is that $C=[0,1]$. For this prove that each side is contained in the other.

Let $a \in C$. To prove that $a \in [0,1]$ assume the contrary and arrive at a contradiction: if $a \notin [0,1]$ then either $a <0$ or $a >1$. In the first case take $\epsilon=-a$ and consider $(a-\epsilon,a+\epsilon)$ and try to get a contradiction to the definition of a cluster point. Similarly take $\epsilon=a-1$ in the second case.

Now prove that every point of $[0,1]$ is a cluster point. You should be able to prove this from the definition.

Answer 2

Following Kavi's advice, here is my 2nd attempt:

Let $C$ be a set of all cluster points of $A$ in $\mathbb{R}$. Let $a \in C$ be a cluster point.

Case 1: $a<0$

Let $\epsilon=-a>0$ be given, then $\exists x \in A=(0,1)$ such that,

$0<|x-a|<\epsilon \implies$ $2a < x < 0$ which is a contradiction since $x \in A$. So $a \not\lt 0$.

Case 2: $a > 1$

Let $\epsilon = a - 1 > 0$ be given, then $\exists x \in A$ such that, $0 < |x-a| < \epsilon \implies$ $1 < x < 2a-1$ which again is a contradiction since $x \in A$. So $a \not\gt 1$.

Case 3: $a \in [0,1]$

Claim: all points in $[0,1]$ are cluster points of $A$ in $\mathbb{R}$.

We'll show for any $\epsilon > 0$ and any $a \in [0,1]$, $\exists x \in A$ such that, $0 < |x-a| < \epsilon$.

Consider the boundary conditions first:

Let $\epsilon>0$ and let $a=0$, then $\exists x \in A$ such that, $0<|x-a|<\epsilon \implies$ $-\epsilon < x < \epsilon$. This holds true for $x \in A=(0,1)$, so $a=0$ is a cluster point.

Let $\epsilon>0$ and let $a=1$, then $\exists x \in A$ such that, $0<|x-a|<\epsilon \implies$ $1-\epsilon < x < 1+\epsilon$. This again holds true for $x \in A=(0,1)$, so $a=1$ is also a cluster point.

Now let $\epsilon>0$ and $0<a<1$, then $\exists x \in A$ such that, $0<|x-a|<\epsilon \implies$ $a-\epsilon < x < a+\epsilon$. This again holds true for $x \in A=(0,1)$, so $0<a<1$ is also a cluster point.

Putting it all together, $C=[0,1]$ is a set of all cluster points of $A$ in $\mathbb{R}$. $\Box$