Calculus – Proof That Every Point of Every Open Set $E \subseteq \mathbb{R}^2$ is a Limit Point of $E$

calculusproof-verification

NOTE: There are some other similar questions, but I wanted to know if my proof is valid. Also note this is a graded assignment.

  1. Is every point of every open set $E⊂{ℝ}^2$ a limit point of E? Answer the same question for closed sets in $E⊂{ℝ}^2$

From Principles of Mathematical Analysis

Proof: By definition, as E is an open set, all points that belong to $E$ are interior points. Therefore there exists a neighborhood for all points of $E$ such that neighborhood $N ⊂ E$. A neighborhood on a plane (in ${ℝ}^2$) is a disk with a radius $r > 0$.

We now have:
For $∀p$, such that $p∈E$, there there exists a neighborhood ${N_p} ⊂ E$ with a radius $r > 0$.

Then there exists a $q$, such that $q∈N_p$ and $q ≠ p$. Let $p$ (an ordered pair) = $(x_1, y_1)$ and $q$ (also an ordered pair) = $(x_2, y_2)$. Therefore, there exists another point, $m$, between $p$ and $q$. $ m = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})$. Therefore p is a limit point. [QED]

Part 2: Counterexample. The set {(0,0)} is closed and (0,0) is not a limit point of $E$.

I am new to this area of mathematics, so I hope my proof isn't a total mess. I don't think it is very good and I would like advice on this proof and whether or not it is valid. I worry it has the correct idea but perhaps it is incomplete (do I need to use induction to show that for all points m, there is another m between it and p?).

Best Answer

In fact, I don't think you ever arrive at the definition of a limit point, so your proof is not complete. In particular, you haven't shown that every neighborhood of $p$ contains an element of $E$.

Here's what I think of as a "cleaner" and completed version of your proof:

Claim: Every point of an open set $E \subset \Bbb R^2$ is a limit point.

Proof:
Consider a point $p = (x_1,y_1) \in E$. By the definition of an open set, there exists an $r>0$ such that $N_r(p) \subset E$. We note that for any $s$ such that $0 < s < r$, the point $q_s = (x_1 + s,y_1)$ satisfies $d(p,q) = s$. So, $q_s \in N_{r}(p) \subset E$.

Thus, for every $\epsilon > 0$, if we select $s$ with $0<s < \min\{\epsilon,r\}$, then we find $q_s \in E$, $d(p,q_s) < \epsilon$, and $q_s \neq p$. It follows that $p$ is a limit point of $E$.

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