Prove that a subset C of $\mathbb R^n$ is closed if and only if it contains all its limit points.

Question

Prove that a subset C of $\mathbb R^n$ is closed if and only if it contains all its limit points.

A closed set is defined by a set of all boundary points

My professor said "We may prove that C is not closed if and only if $C^c$ has a limit point of C. To prove this, it suffices to show that a point x in $C^c$ is a boundary point of C if and only it is a limit point of C." and he left it as an exercise.

I am a beginner of analysis and I want you to check if my proof is okay.

This is how I prove:

If part) Let $x\in C^c$ and x is a limit point of C. Then there is a sequence $\{x_k\}$ of distinct points in C such that $0<\Vert x_k -x \Vert<1/k$. Thus $N'(x;r)\bigcap C$ is infinite. It follows that $N(x;r)\bigcap C\neq\emptyset$. Since $x\in C^c$, we get $N(x;r)\bigcap C^c\neq\emptyset$. Hence x is a boundary point of C.

Only if part) Let $x\in C^c$. Since x is a boundary point of C, there exists $x_k\in C$ such that $x=\lim_{k\to \infty} x_k$. Choose $\{x_{k_j}\}$ where $x_{k_j}\neq x_{k_l}.$ Then $\Vert x- x_{k_j} \Vert > \Vert x- x_{k_{j+1}} \Vert$, so $\{x_{k_j}\}$ converges to x. Thus x is a limit point.

Please tell me if there is wrong or insufficient. Is it okay just to say that "Choose $\{x_{k_j}\}$ where $x_{k_j}\neq x_{k_l}$"?

Answer 1

For topological space, the usual definition of a closed set is that it contains all its limit points. So your statement to prove is in fact a definition.

OK so here, instead, we define a set to be closed if it contains all its boundary points.

I think your only if part can be simplified.

If $x$ is a boundary point of $C$ and $x\in C^c$, then by definition there exists a sequence ${x_k}$ where $x_k \ne x$ for all $k$ and $$\lim_{k\rightarrow\infty}x_k=x$$.

In fact, I think the proof is much simpler if you use, instead of sequence, the following definition of limit point: $x$ is a limit point of $C$ if every neighbourhood of $x$ contains at least one point of $C$ different from $x$ itself.

Then the proof is much simpler:

Let $V_x$ be any neighbourhood of $x$. $$C \text{ is not closed} \iff \exists x \in \partial C \text{ and } x \in C^c$$ $$\iff \exists x'\ne x, x' \in C, x'\in V_x$$ $$\iff x \text{ is a limit point of }C$$