Basic topology: having trouble understanding a couple of things in Baby Rudin

Question

I'm reading Baby Rudin's chapter 2 concerning Euclidean space. First, I find these two statements very confusing to me, which seemingly contradict:

(1). In Example 2.16 there is a statement that
Every subset of a Euclidean space is a metric space.
(2). Beneath Theorem 2.33 there is yet another statement saying that
Every metric space X is an open subset of itself, and is a closed subset of itself.

I understand (1) perfectly well, but can't figure out why (2) is true. For example, an open neighbourhood is a subset to the whole Euclidean space, thus by (1) it is a metric space in its own right, but how can it be both an open and a closed subset to itself? Isn't it an open set itself?

Another thing I have trouble understanding is the relativeness of being a closed set, say, X is closed relative to Y. I simply can't figure out an example in which X is closed relative to Y but not closed to the whole Euclidean space. To me, being closed is equivalent to containing all its limit points, hence if X is closed relative to Y, doesn't it imply that X already contains all its limit points and is thus closed relative to the whole space? Can you provide me with some examples to illustrate this relativeness? (By the way I understand the relativeness of being open. For example, the segment $(0,0)\to(1,0)$ excluding the two ends is open relative to the X axis but surely not open in $\Bbb R^2$. However, it's been difficult for me to understand analogously the relativeness of being closed: say, the "closed" segment (containing the two ends) $(0,0)\to(1,0)$ is closed both in X axis and in $\Bbb R^2$, isn't it?)

Best regards!

Answer 1

$X$ is open because for all $x \in X$, $B_1(x) \subseteq X$.

It can ALSO be closed. In topology, a set is not like a door. It may be open, it may be closed, and it may be neither.

For "$X$ is closed" I think it's best to think of the theorem (or definition, depending on the book) "A set, $S$ is closed if and only if $S^c$ is open". The compliment of $X$ is the empty set which is VACUOUSLY open (for all $x\in \emptyset$, you can say whatever you want).

To give an example of a set that is closed relative to another set, but not to the whole space. Take the subset $(0,1) \subset \Bbb{R}$. Then $(0,1)$ is closed relative to $(0,1)$ but it is not closed relative to $\Bbb{R}$.

Answer 2

A topology on a set A is a collection B (the open sets) of some or all of the subsets of A, subject to certain conditions, one of which is that A and the empty set belong to B. If B is a topology on A, and C is a subset of A, the subspace topology on C, with respect to B, is the family of intersections of C with the members of B. The set A belongs to B,and its intersection with C is C. Now if D is a subset of C, then " D is closed in C" means that any limit point of D WHICH BELONGS TO C is a member of D. For example if A is the real numbers with B being the usual topology, with C being the positive reals , and D the set of reciprocals of positive integers, then D is closed in C . The limit point 0 does not belong to C. The set D is not closed in A.