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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?

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    $\begingroup$ 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 $Y \subset X$? $\endgroup$ – Mathemagical Nov 12 '17 at 4:46
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Nov 12 '17 at 4:54
  • $\begingroup$ 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. $\endgroup$ – fleablood Nov 12 '17 at 5:50
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The answer to your first question is yes.

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 $Y.$

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  • $\begingroup$ But that does not mean that the set G always has to be open. Is there any counterexample for this? $\endgroup$ – 민찬홍 Nov 12 '17 at 5:00
  • $\begingroup$ 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. $\endgroup$ – spaceisdarkgreen Nov 12 '17 at 5:10
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  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.
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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 $G\cap Y=[0,1]$.

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