[Math] Product, Subspace and Order Topology

general-topology

When does product topology , subspace topology and order topology coincide?

Let $ X$ and $Y$ be two ordered sets in their ordered topologies. Let $U$ be a subset of $X$ that is convex in $X$ and $V$ be a subset of $Y$ that is convex in $Y$.

Consider the product topology $X \times Y$.Then will its ordered topology on $U \times V$ be the same as its subspace topology of $X \times Y$?

Best Answer

Background:

Let $X$ and $Y$ be topological spaces with nonempty subsets $A \subset X$ and $B \subset Y$. Now, as you've rightly observed, $A \times B$ can be topologised in two natural ways:

  1. Give $X \times Y$ its product topology. Then, give $A \times B$ the subspace topology it inherits as a subset of $X \times Y$.
  2. Give each of $A$ and $B$ its subspace topology. Then, give $A \times B$ the product topology coming from the topologies on $A$ and $B$.

Fact 1: Both of the above procedures lead to the same topology on $A \times B$.

This fact is standard, and it shouldn't be difficult to find a proof online. It is also not very difficult, so I would encourage you to prove it yourself. Now, it happens that this fact can be refined somewhat with a sort of uniqueness statement.

Fact 2: If $\tau_A$ and $\tau_B$ are topologies on $A$ and $B$ whose product topology $\tau_A \times \tau_B$ (this is notational abuse, the product topology is not really a Cartesian product) equals the topology (2) above, then $\tau_A$ is the subspace topology on $A \subset X$ and $\tau_B$ is the subspace topology on $B \subset Y$.

The reason Fact 2 is true is that, whenever you form a product topological space $X \times Y$, you can recover the topologies on $X$ and $Y$ from the topology on $X \times Y$ by looking at "slices" $\{ (x,y_0) : x \in X\}$ and $\{ (x_0, y) : y \in Y\}$ where $x_0 \in X$ and $y_0 \in Y$ are fixed. These slices turn out to be homeomorphic to $X$ and $Y$ under the natural bijections. So, as a consequence, a product topology on $X \times Y$ is uniquely determined by the input topologies on $X$ and $Y$.

Now we bring in the orders:

Now, in the situation you describe, $X$ and $Y$ are ordered, and equipped with the order topology. You have selected subsets $U \subset X$ and $V \subset Y$. There are now three natural ways to topologise $U \times V$:

  1. Give $X \times Y$ its product topology. Then, give $U \times V$ the subspace topology it inherits as a subset of $X \times Y$.
  2. Give each of $U$ and $V$ its subspace topology. Then, give $U \times V$ the product topology coming from the topologies on $U$ and $V$.
  3. Consider $U$ and $V$ to be ordered using the suborders inherited from $X$ and $Y$. Then, give each of $U$ and $V$ its order topology. Finally, give $U \times V$ the product topology coming from the order topologies on $U$ and $V$.

Now, procedures (1) and (2) always lead to the same topology on $U \times V$ by Fact 1. Meanwhile, by Fact 2, (2) and (3) will lead to the same topology on $U \times V$ if and only if the input topologies on $U$ and $V$ used to form the product are the same. So, the question you should really be asking is:

Question: Let $X$ be an ordered space and $U \subset X$ a subset. There are two natural ways to topologise $U$.

  1. Make $U$ into an ordered space in its own right by viewing it as a suborder of $X$. Then, give $U$ its order topology.
  2. Give $X$ its order topology. Then, topologise $U$ as a subspace of $X$.

When are these two topologies on $U$ the same?

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