# Understanding the Proof of Relative Openness Theorem in Rudin's Principles of Mathematical Analysis

I'm having trouble understanding Rudin's proof for the theorem stating:

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

In particular, I'm having trouble understanding the direction going from right to left. He says:

"Conversely, if $$G$$ is open in $$X$$ and $$E = G \cap Y$$, every $$p$$ in $$E$$ has a neighborhood $$V_p$$ in $$G$$. Then $$V_p \cap Y$$ is a subset of $$E$$, so that $$E$$ is open relative to $$Y$$."

I'm having trouble understanding how to justify the last part where he concludes that the intersection $$V_p \cap Y$$ is a subset of $$E$$.

• $V_p$ is in $G$, so $V_p\cap Y$ is in $G\cap Y$, which is exactly $E$.
– MPW
Commented Jun 17 at 17:58
• Think about how to prove $[0,1)$ is open in $[0,\infty),$ with the usual absolute value metric/topology. Commented Jun 18 at 0:16

This may be an approach to consider (expanding somewhat on the comment by @MPW). Focusing on the "right-to-left" part of the proof:

Suppose $$G$$ is open in $$X.$$ This means that every point of $$G$$ is an interior point of $$G.$$ This in turn means that there is a neighborhood $$V_{p}$$ such that $$V_{p} \subset G.$$ This is from Definition 2.18 (f) & (e) in Rudin Principles of Mathematical Analysis 3rd Ed.

Now suppose that $$E=G \cap Y.$$ This means that every point $$p \in E$$ is also a point $$p \in G$$ by definition of intersection. And from the paragraph above, every point of $$E$$ has a neighborhood $$V_{p} \subset G.$$

First, it is helpful to prove that if $$A \subset G$$ and $$G \cap Y \subset E$$, then $$A \cap Y \subset E$$, where $$A$$ is an arbitrary set.

We note that $$A \cup G = G$$, so we can write $$\left(A \cup G\right) \cap Y \subset E$$. Then we have $$\left(A \cup G\right) \cap Y= \left(A \cap Y\right) \cup \left(G \cap Y \right) \subset E.$$ This is justified by the Remarks 2.11 in Rudin.

Now, we have $$\left(A \cap Y\right) \cup \left(G \cap Y \right) \subset E \implies \left(A \cap Y\right) \cup E \subset E.$$ Then we have $$A \cap Y \subset \left(A \cap Y\right) \cup E \subset E$$ from Rudin Remarks 2.11 (11). Thus $$A \cap Y \subset E.$$

Now, substituting $$V_{p}$$ for $$A$$, we have $$V_{p} \cap Y \subset E.$$

• How is G not just E (i.e. E=G)? Doesnt every neighborhood of every point in E also contain every point in E? And every other point in the neighborhood is also in E by the definition of relative openness? I think this is what I'm struggling to wrap my head around. This seems really redundant. Dont we already know that p is in Y (since E is subset of Y) and that p must also be in some neighborhood V_p, so obviously p must be in the intersection of V_p and Y? What am I missing? Commented Jun 17 at 19:25
• Perhaps an example might be useful: Let $X = \mathbb{R}, Y= \left(0,1 \right),G= \left(\frac{1}{2},2\right).$ Then $E= Y \cap G=\left(\frac{1}{2}, 1 \right).$ Thus $E \ne G.$ Commented Jun 17 at 19:40
• ahh I think I see what my confusion was. The definition of relative openness says that the members of the subset E of Y are also members of E whenever those points are also in Y, but there could be points inside some neighborhood V_p that aren't in Y. Commented Jun 17 at 22:19

The subspace topology is usually defined as the statement given in the proposition. Rudin however defines "$$E$$ open relative to $$Y$$" in the context of metric spaces. The definition provided in the preceding paragraph is

$$E$$ is open relative to $$Y$$ if to each $$p \in E$$ there is associated an $$r > 0$$ such that $$q \in E$$ whenever $$d (p, q) < r$$ and $$q \in Y$$.

You can understand the converse proof in this way. The point $$p \in E$$ is taken to be arbitrary. Because $$G \subseteq X$$ is open, and $$p \in E \implies p \in G$$, then there is some basis element $$V_p$$ of $$p$$ in $$X$$ such that $$V_p \subseteq G$$. In a metric space, the basis elements are the open balls, so we can take $$V_p = B_d (p, \varepsilon) = \{ q \in X | d (p, q) < \varepsilon \}$$ for some $$\varepsilon > 0$$. Looking at Rudin's definition, we can simply take $$r = \varepsilon$$ to complete the proof. Indeed, take any $$q \in Y$$ that satisfies $$d (p, q) < \varepsilon$$ (note that this is equivalent to the statement $$q \in Y \cap V_p$$). Then since $$V_p \subseteq G$$, this shows that $$q \in Y \cap G = E$$, which is what is needed.

First off, it's important to understand what's being proved here is precisely that the "subspace topology" is $$\textit{equal}$$ to Rudin's definition of being "relatively open" (e.g. $$(-1, 1)$$ is open in $$\mathbb{R}$$ but not in $$\mathbb{R}^2$$).

In other words, a set will be open under one definition iff it will also be open under the other definition. With that in mind, let's get started.

Say $$E \subset Y \subset X$$. We want to prove $$E$$ is open relative to $$Y$$ iff $$E = Y \cap G$$ for some $$G \overset{\text{open}}{\subset} X$$.

$$\implies)$$ If $$E$$ is open relative to $$Y$$, then for every $$p \in E$$, there exists some subset $$V_p \overset{\text{open}}{\subset} X$$ containing $$p$$ so that if $$q \in V_p$$ and $$q \in Y$$ then $$q \in E$$. Call $$W_p \equiv V_p \cap Y$$. By what we just said, $$W_p \subset E$$. So if you think about it, $$W_p$$ is really functioning like an open set of $$Y$$ contained in $$E$$ which contains $$p$$. Because we can do this for every single $$p \in E$$, $$G \equiv \cup_{p \in E} V_p$$ is an open set of $$X$$ such that $$G \cap Y = E$$ because if a point is in $$G$$, it is in some $$V_p$$ so if it is in $$Y$$ then it must also be in $$E$$ and clearly if a point is in $$E$$, it will be in $$G$$ and $$Y$$.

$$\Longleftarrow)$$ Say $$E = Y \cap G$$ for some open subset $$G$$ of $$X$$. Pick some arbitrary $$p \in E$$. To show $$E$$ is open relative to $$Y$$, we want to find a neighborhood of $$p$$ that is open in $$X$$ such that if we intersect that nbhd with $$Y$$, it will only contain elements of $$E$$. Well, $$E \subset G$$ and $$G$$ is open so there is some open neighborhood of $$X$$, say $$V_p$$, which contains $$p$$ and is inside $$G$$. $$p \in V_p \cap Y \subset G \cap Y = E$$ so we are done. :)