# Is it true that the definition of an open subset in a metric space is different from the combination of the definitions of subsets and opens sets?

I have a question concerning the equivalence of two definitions of open subsets (in metric spaces). To avoid confusion, I will state the two definitions and then ask my questions.

The definition I found is: Given a metric space $X$ and let $S \subset Y \subset X$. $S$ is an open subset of $Y$ if $\forall p \in S, \exists r > 0 \ | \ d(p,q)<r, q \in X \rightarrow q \in S$. In contrast, I simply defined $S$ to be an open subset of $Y$ if it is open and a subset of $Y$. The two definition however seem (at least to me) to be quite different because for the set $S$ to be open one only considers whether $S$ consists solely of interior points or not. The first definition however checks the existence of a neighbourhood of $p$ such that all points which are element of the largest set $X$ are also part of the smallest set $S$.

I have the following questions in particular:

• Is it really true that the definition of a open subset is not a combination of two definitions of an open set and a subset?
• The part "$q \in X$" of the first definition appears a bit strange to me. Is it correct that we do not considers points in Y but instead of the larger set $X$? What is then an open subset if one considers two sets instead of three?

I am looking forward for your replies.

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A simple example is $X=\mathbb R^2$ and $Y=\mathbb R\times\{0\}$. Then the standard definition makes the open sets of $Y$ the same as the open sets of $\mathbb R$, while your definition would have no open subsets of $Y$.
If $Y$ is open in $X$, then the two definitions coincide.
It turns out that $S$ is open in $Y$ (under the usual definition) if and only if $S=T\cap Y$ for some open subset $T\subseteq X$.
@fabian: When we talk about being open in a subset $Y$ of a given space, we do indeed mean open relative to $Y$. When we just say "open," we mean open in the given space. – Cameron Buie Aug 1 '13 at 15:13