# Questions concerning Theorem 2.30 of Baby Rudin

First question I have is the following:

Is [0, 1] open relative to [0, 1]?

It seems open to me because for x in (0, 1), x is definitely an interior point of [0, 1] and for x=0, 1 there is a neighborhood centered at each point which is completely contained in [0, 1] because x>1 or x<0 are not in our attention.

If the answer is positive to the previous question, then I have this further question.

Theorem 2.30 of Baby Rudin is stated as follows.

Suppose $Y\subset X$. A subset E of Y is open relative to Y if and only if $E=Y \cap G$ for some open subset G of X.

If we consider a specific case like Y=[0, 1], X=$\mathbb {R}$ and E=Y, then G can be Y, so should G always be an open set?

My last question is the following. Is the purpose of implementing the new set G in the above Theorem to 'delete' all the points in X which are not in the sets containing the set E? Thus when G is intersected with Y, the result is the set E?

• Several typos (non trivial ones). Please go through them carefully and edit. Otherwise, it is hard to understand what you are saying. $Y \in X$ or $Y \subset X$? Nov 12, 2017 at 4:46
• For the 2nd Q, G must be open in X. But Y is not necessarily open in X. For the first Q, take the def'n in the 2nd Q and consider the case G=X. Nov 12, 2017 at 4:54
• 1) as a space, every set is open relative to itself (because the space itself is always open). 2) $G\ne [0,1]$ as $[0,1]$ is not open in $\mathbb R$. But $[0,1]\subset (-1,2)$ and $[0,1]=[0,1]\cap (-1,2)$ so everything is cool and frody. Nov 12, 2017 at 5:50

As for the second question, the definition is correct as stated. Just because $Y$ is the intersection of $Y$ and a non-open subset of $X$ ($Y$ in this case) does not mean there isn't an open subset $G$ of $X$ such that $Y$ is the intersection of $G$ and $Y$ (any open superset of $Y$ will do here.)

For the third question, I'm not sure I understand exactly, but it seems this is at least on the right track. We are restricting to $Y$ and part of the idea is to guarantee $Y$ is open in $Y.$

• But that does not mean that the set G always has to be open. Is there any counterexample for this?
– 민찬홍
Nov 12, 2017 at 5:00
• I don't know what you mean by "always has to be open". In order for the set $E$ to be open relative to $Y$ there needs to be some set $G$ that is open relative to $X$ with $G\cap Y = E.$ This does not mean that every set such that $G\cap Y =E$ is open relative to $X.$ There may be many sets $G$ such that $G\cap Y = E,$ some open, some not open relative to $X.$ If this is the misconception you're having then yes, $E = [0,1],$ $Y=[0,1],$ $X=\mathbb R,$ $G=[0,1]$ is a counterexample showing this is a misconception. Nov 12, 2017 at 5:10
1. Yes. The whole space is always a member of a topology (that is, an open set). So $[0,1]$ is open in any topology on $[0,1]$.
2. G needs to be open in the topology on X.
3. You can say that the idea is indeed to “delete” (disregard) points not in Y.

It might be helpful to keep the definition of a topological space in mind here. Remember that a topological space is a set $X$, together with a collection of subsets which satisfy some properties.

To say that a set $Y$ is open relative to another set $X$ is to say that $Y$ is among the collection of subsets which form a topology on $X$.

So your first question, "is $[0,1]$ open relative to $[0,1]$?" The answer is yes because (as part of the axioms of a topology) the whole space $[0,1]$ must be an open subset of $[0,1]$ (this is true for any topological space).

For your second question, what is the theorem saying? It states that for a fixed subset $Y\subset X$. A subset $E\subset Y$ is open (in the space $Y$!) if there is a $G\subset X$ open (in the space $X$!) such that $E=G\cap Y$. So in your specific example, $Y=[0,1]$, $X=\mathbb{R}$ and $E=[0,1]$, you cannot choose $G=[0,1]$, because that is not an open subset of $\mathbb{R}$. However, it is open in $Y=[0,1]$ because $G=\mathbb{R}$ is open in $\mathbb{R}$ and $G\cap Y=[0,1]$.