4
$\begingroup$

Lemma: $A\subset\mathbb{R}, z=\inf(A)$, and $z \notin A$, then $z$ is an accumulation point of $A$.

Here is my proof: Let $e>0$ be given. Consider $N(z,e)$. Note that $N(z,e)=(z-e,z+e)$.

Since $z=\inf(A)$, $z+e$ cannot be a lower bound of set $A$. Hence there exists a $t \in A$, such that $z-e

$N$ represents the Neighborhood of $z$.

Any feedback or corrections would be very helpful thank you.

  • 0
    Don't you attend to edit the question?! – 2017-02-28
  • 0
    I have two comments. (1). Your proof is correct. (2). Please format your questions properly and write all math symbols within dollar signs so that other community members have more time to answer the questions rather to make them more readable. – 2017-02-28
  • 1
    You prove that $N\cap A\neq\varnothing$ for every neighborhood $N$ of $A$, or equivalently that $z$ is an element of the closure of $A$. But is that enough for $z$ to be an accumulation point of $A$? Where did you use $z\notin A$? – 2017-02-28
  • 0
    How would I prove that z does not belong to A? @drhab – 2017-02-28
  • 0
    You don't need a proof of that. It is given unto you. You must use this fact to complete the proof that $z$ is an accumulation point of $A$. You have shown allready that $z\notin\bar A$. Now use the fact that every element of $\bar A\setminus A$ must be an accumulation point of $A$. – 2017-02-28

1 Answers 1

2

Two relatively minor points:you get $a \in A$ where $z \le a is already a lower bound. Where you use that $z \notin A$ is the fact that $z < a$ must hold (otherwise we would have that $z = \min(A)$ and this could be isolated: $\{0\} \cup[1,2]$ is an example where this happens). So we have $a \in (A\setminus \{z\})\cap N$ for every neighbourhood $N$ of $z$. This makes it a limit point in my terminology (and as the reals are $T_1$, it is also an ($\omega$-)accumulation point: every $N$ intersects $A$ in an infinite set.