Openness in $\Bbb R^2$ In $\Bbb R^2$


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*Is$\;L_n:= \{(x,{x\over n})\in \Bbb R^2\mid \; x\in [0,1]\},\; n\in \Bbb N$ open?

*Is $ L:= \bigcup_{n=1}^\infty L_n \setminus \{(1,0)\} $ open?
 A: The answer is no for both sets.
Lets recall that a subset $M$ of $\mathbb{R}$ is open (in the standard topology) if for each $x\in M$ there is a $\epsilon>0$ such that $B_\epsilon(x) \subseteq M$.


*

*Lets regard $L_n := \{(x,\frac{x}{n})\in \mathbb{R}^2 \mid \; x\in [0,1]\}$:
Take the point $(1,\frac{1}{n})$ and note that for every $\epsilon > 0$ the point $(1+\frac{\epsilon}{2},\frac{1}{n})$ is in $B_\epsilon\big((1,\frac{1}{n})\big)$ but obviously not in $L_n$. So there is no $\epsilon>0$ such that $B_\epsilon\big((1,\frac{1}{n})\big)\subseteq L_n$.
So $L_n$ can't be open.

*The reason is basically the same:
Note that every $(x_1,x_2)\in L$ satisfies $x_1 \leq 1$. So again $(1+\frac{\epsilon}{2},1) \notin L$ but in $B_\epsilon\big((1,1)\big)$ for every $\epsilon>0$ and $(1,1)$ is an element of $L$.
You regard the sets $L_r' := \{(x,\frac{x}{n})\in \mathbb{R}^2 \mid \; x\in (0,1)\}$ for $r>1$. The sets $L_r'$ are still not open but the union
$$L' := \bigcup_{r\in\mathbb{R}, r>1} L_r'$$
is open. The set $L$ is the triangular of the points $(0,0)$, $(0,1)$ and $(1,1)$ without its border lines.
A: Here is another approach:
A sometimes useful trick is to note that if $f$ is a non zero linear functional (takes values in $\mathbb{R}$) and $U$ is open, then
$f(U)$ is open (in $\mathbb{R}$).
For 1, note that $f_n:\mathbb{R}^2 \to \mathbb{R}$ defined by $f_n(x) = x_1 - n x_2$ is linear, non zero and constant on $L_n$, hence $L_n$ is not open.
A related trick that relies on the inverse function theorem is
to note that if $f$ is a differentiable function such that
$f'(x_0) \neq 0$ at some point $x_0$, and $U$ is an open set containing $x_0$, then $f(U)$ contains some open set $V$ that contains $f(x_0)$.
To apply this trick, note that if $L$ was open, then the
set $L'=L \cap \{ x | x_1 > 0 \}$ would also be open.
The function $f: L' \to \mathbb{R}$ defined by $f(x) = {x_2 \over x_1}$ (cf. the angle of the point $x$) is differentiable, and $f'(x) \neq 0$ for $x \in L'$.
However, we see that $f(L') = \{0\} \cup \{{1 \over n}\}_n$ which
has an empty interior. It follows that $L'$ is not open,
hence $L$ is not open.
