Let $(X,d)$ be a metric space.
Let $A\subset X$.
Verify if the following propositions are true or false.
Proposition 1 $$B\subset A \text{ is closed in A}\Longleftrightarrow B=C \cap A \,\,\,\,\,(C\text{ open in }E)$$
Proposition 2 $$B\subset A \text{ is open in A}\Longleftrightarrow B=O \cap A \,\,\,\,\,(O\text{ open in }E)$$
I am fairly certain Proposition 1 is true, however I am not so certain about Proposition 2.
I feel it is $O\cap A$ insead of $O \cup A$.
Any verification is appreciated.
I cannot remember which textbook I extracted these from.
Propostion 2: (properly restated as) "if $B \subset A$ then $B$ is open in $A \iff $ there is a set $C$ that is open in $X$ so that $B = C \cap A$" is simply the definition of open in a subspace $A \subset X$.
So it is true because it is a definition.
(EDIT: ERK ... not entirely true. But close. For Topological spaces in general, this is a definition. For a metric space this is a proposition but easy to prove.)
(In a metric space $X$ a *neighborhood, $B_r(x)$ is defined $\{y\in X|d(x,y) < r\}$. So if $A \subset X$ is a subspace with the same metric then if $C_r(x) = \{y\in A|d(x,y) < r\}$ is a neighborhood in space $A$ of a point $x\in A$. Then $C_r(x) \subset B_r(x) = \{y\in X|d(x,y) < r\}$, a neighborhood *in space $X$.)
(Thus everything we can conclude about neighborhoods in $X$ will apply to the "subneighborhoods" restricted to $A$. Hence the properties of whether sets are open and closed will be "inherited" from $X$.)
Proposition 1: is simply obviously false. If you take an open set $C$ and intersect it with $A$ there is utterly no reason that that $B = C \cap A$ will be closed when $C$ was open.
It's easy to come up with counter examples. Let $X = \mathbb R$. $A = [0,\infty)$, $C = (1,2)$. Then $B = C \cap A= (1,2)$ is certainly not closed it $A$.