Topology of $\mathbb R^n$ question I have the following problem:
$\newcommand{Rn}{\mathbb R^n}$
$\newcommand{U}{\bigcup}$
$\newcommand{a}{{a \in A}}$
$\newcommand{b}{{b \in B}}$
$\newcommand{es}{\emptyset}$
$\newcommand{ss}{\subset}$
$\newcommand{sm}{\setminus}$
$\newcommand{d}{\partial}$
$\newcommand{e}{\varepsilon}$


  Let $X \ss \Rn$, and let $A$ and $B$ be disjoint open subsets of $X$ with the subspace topology. Show that there exist disjoint open subsets $U$ and $V$ of $\Rn$ such that $U \cap X = A$ and $V \cap X = B$.


My partial solution is as follows:

Since $A$ is open in $X$, for every $\a$ there exists $\e_a &gt; 0$ such that $A = \U_\a \beta_X(a, \e_a)$, where $\beta_X(x, \e)$ is the open ball relative to $X$, centered at $x$, of radius $\e$. Now let $P = \U_\a \beta(a, \e_a)$. Clearly, $P$ is open in $\Rn$ and $P \cap X = A$. Construct $Q$ from $B$ and $X$ in a similar way.

Clearly, $P \cap Q \cap X = A \cap Q = B \cap P = A \cap B = \es$. However, $P \cap Q$ isn't necessarily empty. It is merely guaranteed that their intersection lies outside of $X$, and hence outside of $A$ and $B$. That's why I haven't called them $U$ and $V$ in the first place.

Now, I want to show that $U = P \sm \bar Q$ and $V = Q \sm \bar P$ satisfy the given requirement. Clearly, $U$ and $V$ are both open in $\Rn$, and $U \cap X = (P \sm \bar Q) \cap X = (P \cap X) \sm (Q \cup \d Q) = A \sm \d Q$ (and similarly for $B$ and $V$). I just need to show that $A \cap \d Q = \es$. How do I do this? Or, is my idea completely wrong?
The proposition is false if $\mathbb{R}^{n}$ is replaced by a general
topological space (or the metric topology is replaced by an arbitrary
topology).

For example. Let $\mathbb{R}$ be equipped with the topology $\{\emptyset,\mathbb{R},\{1\}^{c},\{2\}^{c},\{1,2\}^{c}\}$
and let $X=\{1,2\}\subseteq\mathbb{R}$. Let $A=\{1\}$ and $B=\{2\}$,
Note that $A$ and $B$ are open subsets of $X$ because $A=\{2\}^{c}\cap X$
and $B=\{1\}^{c}\cap X$. Clearly $A\cap B=\emptyset$. However, we
cannot find disjoint open sets $U,V$ of $\mathbb{R}$ such that $A=U\cap X$
and $B=V\cap X$.

Therefore, to prove the proposition (if it is true), one must use
properties of metric topology on $\mathbb{R}^{n}$. Any proof that just uses set-theoretic construction on general setting (without anything specific to the metric topology) must be invalid.
Your quoted statement seems to be true by definition? If $A \in X$ is open, then there is an open $U \in \mathbb{R}$ so that $A =U\cap X$. See subspace topology. Or otherwise, you shall define openness in subspace another way.

It remains to show $U \cap V =\varnothing$. If not, then remove the common part of $U$ and $V$. Indeed, if we define $U' =U -\overline{U \cap V}$, then $U'$ is open. Similarly $V' =V -\overline{U \cap V}$, and $V'$ is open. Since $U \cap V \cap X =\varnothing$, we see still $A =U' \cap X$ and $B =V' \cap X$.