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This goes for equipping any set with any topology. Suppose that I want to equip a subset $A \subseteq X$ where $X$ is a topological space, with the subspace topology. I know that by definition of the subspace topology that $\tau_A = \{U \cap A | U \in \tau\}$. But what does this do to $A$?

Joey
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    Consider the plane $P={(x, y, z) \in \mathbb{R}^3 \mid z = 0}$ in $\mathbb{R}^3$. How do you define an open set on $P$? One way is to find open sets in $\mathbb{R}^3$ then intersect them with $P$. – Jacky Chong Feb 20 '21 at 05:41
  • We know that $\Bbb R^2$ has a topology. But we also want to call subsets of the plane "connected" or "compact" etc. For this we need a topology on that subset to consider it a space in its own right, so those properties are defined for it too. This is what the subspace topology achieves. – Henno Brandsma Feb 20 '21 at 08:41

3 Answers3

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A topology $\tau$ on a set $X$ is merely a collection of subsets of $X$ that conforms to certain rules (rules which have been chosen for their utility with respect to proving results elsewhere in mathematics$^\dagger$).

Suppose we have a set $X$ and a topology $\tau \subset \mathcal{P}(X)$. Then, for any subset $Y \subset X$, there always exists, whether explicitly acknowledged or not, what is called the subspace topology, $\tau_{_Y}$, on that subset, which is always defined with respect to an original topology, and whose open sets are, namely, where $Y$ intersects with each of the original open sets. This is to say: $U \cap Y$ is an element of $\tau_{_Y}$ whenever $U \subset X$ is an element of $\tau$.

$\underline{\textbf{Exercise}}$: Verify that $\tau_{_Y}$ as we have defined it is indeed a legitimate topology on $Y$, i.e. that it conforms to the rules a collection of subsets must follow to be a topology.

But the existence of $\tau_{_Y}$ doesn't "verb" anything to $Y$, which still exists independently as itself.

$^\dagger$If we know a given object is a particular kind of topological space, then everything we have proven about that kind of topological space can be automatically applied to that object. This is relevant because it turns out that a number of objects one studies in mathematics are topological spaces in disguise (with some overlying structure) such as your metric spaces from analysis. Cf. my post here along these lines; the motivation for why the topological axioms have been chosen as such is often overlooked at least at the start of the intro course, and I feel that if the motivation were properly addressed from the start, everything would become clear faster.

The idea of a subspace topology provides the rigorous framework upon which statements like this one from calculus can be made:

If a function $f:\mathbb{R} \to \mathbb{R}$ is continuous, then so too is the restriction of that function to a particular interval, say to $[0,1]$.

Here, it is implied that $[0,1]$ is inheriting the subspace topology from $\mathbb{R}$, which is endowed with the so-called Euclidean topology.

Kaj Hansen
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  • So the subspace topology on a set X conforms to the rules of the subspace topology itself. So the set X will follow the $\tau_X = {U \cap X | U \in \tau }$? Is it wrong to say that $X \in \tau_X$? – Joey Feb 20 '21 at 06:21
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    All of that is correct, except for some notational awkwardness just because you forgot the symbol "$X$" was being used for the original set; find-and-replace everywhere you typed $X$ with $Y$ and it's all good & consistent with the notation from the main text of my post. – Kaj Hansen Feb 20 '21 at 06:34
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    But yes, you always have $X \in \tau$, and whenever someone gives you a purported definition for a "topology" you should make sure that is the case, as indeed the whole set being open is one of the stipulations for a topology at all. And so you also have $Y \in \tau_Y$ for any subset $Y \subset X$. – Kaj Hansen Feb 20 '21 at 06:34
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    Topology was originally called Analysis Situs (Latin situs=place, position, site) . – DanielWainfleet Feb 20 '21 at 16:10
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To talk about a topological space, you need a set and a topology on that set.

If you are given a topological space $X$, you're actually given a pair $(X, \mathcal T_X).$

Now, if you're given a subset $A \subseteq X$, you want to get a topological space out of it. To say that "$A$ is equipped with the subspace topology", we mean that we're looking at the topological space $(A, \mathcal T_A)$, where $\mathcal T_A$ is as you described.

Note that this mainly stems from the abuse of notation we do when we say something like "Let $X$ be a topological space", instead of using $(X, \mathcal T_X)$.

Aryaman Maithani
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  • Just to clarify, the topological space $(A, \tau_A)$ will follow the "rules" set by the topology on $A$ or am I misunderstanding whats going on – Joey Feb 20 '21 at 06:13
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    @Joey Yes, $\tau_A$, as defined in your question, will satisfy the axioms for a topology, and so $(A, \tau_A)$ is a topological space (and has all the usual properties of topological spaces). – Theo Bendit Feb 20 '21 at 06:28
  • @Joey, it is actually a good exercise, given how the subspace topology has been defined, to prove for yourself that the collection of subsets $\tau_Y$ indeed obeys the axioms of a topology. Nothing especially difficult, but a good exercise nonetheless. – Kaj Hansen Feb 20 '21 at 08:27
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Nothing is altered about the set $A$ itself. Instead, the language "to equip with a topology" means to spawn a topological space $(A, T_A)$ with the topology $T_A$ being the topology in question. The topological space $(A, T_A)$ and the set $A$ are not the same thing. $A$ forms a part of it, but it is not identical to $A$, nor does it alter the definition of $A$.

How much differentiation there is between these two things depends on how rich our formalizing foundations are. With ordinary, "ZFC" set theory, $(A,\ T_A)$ is just an ordered pair of sets, which in turn is the set $\{ \{ A\}, \{ A, T_A \} \}$. Note this is not equal to $A$, nor does it have $A$ as a direct member. A richer system of foundations, though, would incorporate a typed language and say that $(A, T_A)$ is an object of a different "data type", the type of topological spaces, and thus could not be interconverted to a set or other type of ordered pair unless explicit type casting rules are defined.

The_Sympathizer
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