My real-analysis text gave the following defintion:
Let U be a subset of E. U is open relative to E if
for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \cap E \subset U$.

Although the idea that U is open in $\mathbb R$ follows the definition,
I normally do not think about the intersection of $N_\epsilon(t) \cap \mathbb R$.
U is open if for $\forall t \in U$, $\exists \epsilon$ such that $N_\epsilon(t) \subset U$; every t is an interior point of U.
Intuitively, each point, t, is contained in a "bubble".

So, what is the significance of specifying the intersection?
Apparently, I can't think of a relative open set in terms like interior points and "bubbles".

  • $\begingroup$ $\{0\}$ is open in $\{0,1\}$. $\endgroup$
    – tomasz
    May 29 '13 at 17:40
  • $\begingroup$ As tomasz said $\{0\}$ is open in $\{0,1\}$ but is it open in $\mathbb{R}$ too? $\endgroup$
    – Xena
    May 29 '13 at 17:51
  • $\begingroup$ what does the notation N here mean? $\endgroup$ Oct 12 '20 at 7:02

A relative open set is essentially the restriction of an open set to a subset. For example, open sets in $\mathbb R$ are such that each point in the set is contained in an open interval or "bubble" as you say. Given this idea of openness, how would we define an open subset of the closed interval $[0,1]$? It seems clear that something like $(\frac{1}{4}, \frac{3}{4})$ should be open but what about $[0,\frac{1}{2})$? This set is not open in $\mathbb R$ because any open interval around $0$ contains a negative number which is not in $[0,\frac{1}{2})$. However, there are no negative numbers in $[0,1]$ so this shouldn't preclude this set from being open relative to $[0,1]$. From inside $[0,1]$, $[0,\frac{1}{2})$ looks like an open neighborhood or your "bubble".


As you said, a subset $U$ of $\mathbb{R}$ is open, if for every point $x$ of $U$, there is some $\epsilon>0$ such that $N_\epsilon(x) \subset U$, which means "every point of $\mathbb{R}$ that is closer to $x$ than $\epsilon$ lies in $U$".

Now we want to talk more generally about open subsets of $E$, where $E \subset \mathbb{R}$, but using the same distance function; i.e. keeping as much the same as possible.

So we have $U \subset E$ and we want to define what it means to be open in $E$. The easiest way is to say: for every $x \in U$ there is some $\epsilon > 0$ such that "every point of $E$ that is closer to $x$ than $\epsilon$ lies in $U$". The statement between quotes is just $E \cap N_\epsilon(x) \subset U$, as your text's definition. It makes no sense to consider points not in $E$; we consider $E$ to be the new "universe of discourse", as we only talk about its subsets, no longer about all subsets of $\mathbb{R}$.

  • $\begingroup$ TY. I think your description that "every point of E that is closer to x ..." is much clearer than my description of a bubble. You are adding the condition that the point in consideration is an element of E; $E \cap N_\epsilon (x) \subset U$. Thanks to all who took the time to answer. $\endgroup$
    – Andy Tam
    May 30 '13 at 3:48
  • $\begingroup$ what does N mean here? $\endgroup$ Oct 12 '20 at 7:26
  • $\begingroup$ @JosephRock $N_\varepsilon(x)=\{y: d(x,y) < \varepsilon\}$ is the $\varepsilon$-neighbourhood of $x$. $B(x,\varepsilon)$ or variations thereof are also used. I followed the OP's notation. $\endgroup$ Oct 12 '20 at 7:33

The idea is that you are giving the subset $E$ of $\mathbb{R}$ a topology. A topology on a set is just a collection of subsets $\mathcal{T}$ that satisfy the following three conditions:

  1. $\emptyset$ and $E$ are in $\mathcal{T}$.

  2. $\mathcal{T}$ is closed under finite intersections (i.e. if $U_1,\ldots U_n$ are in $\mathcal{T}$, then so is $U_1\cap\cdots\cap U_n$) .

  3. $\mathcal{T}$ is closed under arbitrary unions.

call the elements of $\mathcal{T}$ open sets of $E$. In $\mathbb{R}$ you know what sets are open. When you take a set together with a topology (called a topological space), you can determine what it means to be open in a space that is not something comfortable like $\mathbb{R}$.

In your example, you are declaring which subsets of $E$ are open. In fact, the topology you are giving $E$ in this context is what is called the subspace topology. You defined what it means for a set to be relatively open in $E$. A nice exercise is to show that the collection of relatively open subsets of $E$ satisfy the properties above.


Given a topological space $(X,\tau_X)$ and an arbitrary set $Y \subseteq X$ we want to endow $Y$ with the structure of a topological space itself. One natural thing to do is to "restrict" the topology $\tau_X$ on $X$ to topology $\tau_Y$ on $Y$. We have to make sure that:

  • $Y, \emptyset \in \tau_Y$
  • $\forall \mathcal U \subseteq \tau_Y: \ \bigcup \mathcal U \in \tau_Y$
  • $\forall U,V \in \tau_Y: \ U \cap Y \in \tau_Y$

We can achieve this by taking the "relative topology" as $\tau_Y$ defined by

$\tau_Y = \{ Y \cap U \in \mathcal P(X) \mid U \in \tau_X \}$.

You might want to check that this indeed gives a topological space $(Y,\tau_Y)$ and that it coincides with your description if we take $X = \mathbb{R}$ and $Y = E$. If you know what a "metric space" is, you might want to think about this case in a more general context...


Here is another example which should help form a mental picture of the notion:

In $\mathbb{R}^3$, a "floating square" without its contour, such as the set $(0,1) \times (0,1) \times \{ 0 \}$ is neither open nor closed. However it is open relative to its affine hull, which is the plane $(x,y,0)$.

That's because for every point in the set there is a ball whose intersection with the plane is included in the set. The intersection between the ball and the plane can be aptly called a relative neighborhood.


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