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 $E$.

  • $\begingroup$ $V_p$ is in $G$, so $V_p\cap Y$ is in $G\cap Y$, which is exactly $E$. $\endgroup$
    – MPW
    Commented Jun 17 at 17:58
  • $\begingroup$ Think about how to prove $[0,1)$ is open in $[0,\infty),$ with the usual absolute value metric/topology. $\endgroup$
    – Steen82
    Commented Jun 18 at 0:16

3 Answers 3


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 $V_{p} \cap Y \subset E.$

  • $\begingroup$ 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? $\endgroup$
    – user8083
    Commented Jun 17 at 19:25
  • $\begingroup$ 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 $E \ne G.$ $\endgroup$
    – ad2004
    Commented Jun 17 at 19:40
  • $\begingroup$ 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. $\endgroup$
    – user8083
    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. :)


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