Is $(-\infty, 0]$ closed in $\mathbb{R}$? Using Rudin's definition that a closed set contains all of its limit points, can I say that the interval  $(-\infty, 0]$ is closed in $\mathbb{R}$ (using the regular metric $d(x,y) = |x-y|$ for $x,y \in \mathbb{R}$)?
 A: Let $A=(-\infty,0]$, so $A^c=(0,\infty)$, which is clearly open. Hence $A$ is closed.

To show that this definition of a closed set is equivalent to your definition:
Here is the proof that $B$ closed if and only if $B^c$ open.
Suppose $B$ is closed (with your definition of closed). Then choose $x\in B^c$. That is, you choose $x\notin B$. Since $x\notin B$, we know that $x$ is not  a limit point. Since it's not a limit point, then we know that there exists some a $\epsilon$-neighborhood for which $N_{\epsilon}(x)\cap B=\emptyset$ where $N_{\epsilon}(x)\subset B^c$. Hence we have that $x$ is an interior point of $B^c$ and consequently $B^c$ is open.
For the reverse direction:
Suppose that $B^c$ is open, so every point in $B^c$ is an interior point. Let $x$ be a limit point of $B$. Then $\forall\epsilon>0$ we have $N_{\epsilon}(x)\cap B\neq \emptyset$, so $x$ is not an interior point of $B^c$. Hence $x\in B$ and so every limit point of $B$ is a point in $B$ and hence $B$ is closed.
Hence $B$ is closed if and only if its complement $B^c$ is open.
A: Assume for the contrary that $(-\infty,0]$ does not contain all its limit points. Let $a$ be sucha limit point. Then $a>0$. But no sequence of points in $(-\infty,0]$ can ever get closer to $a$ than by a distance of $a$.
A: yes...because $(-\infty, 0)$ is open in $\mathbb{R}$ since every point is an interior point...so its complement is closed which is $[0, \infty)$
A: Yes. Assume $x\in\mathbb{R}$ is a limit point of $A=(-\infty,0]$
but $x\not\in A$ then $x>0$ and $(\frac{x}{2},2x)$ is an open neighborhood
of $x$ that does not contain any element of $A$.
A: Yes. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}$
