0

I'm working on generating a counter example that shows projections from $X \times Y$ do not always map closed sets to closed sets. I'm working backwards, that is, from the decompositions to the product space. My attempt is to see if I might generate a closed product space whose projection maps to clopen sets.

I don't know if my example is on the right track, but in any case I'm trying to gain some intuition into the kind of product space generated by these intervals. For example, in the lower left hand corner of the product, I believe the point in $\mathbb{E}^2$ looks something like $(0, 1/n)$. Is this right? If so, is there an "opening" in the product space at $(0,0)$? Is this even the right way to think about this, or is my intuition misguided?

innumeratus
  • 73

2 Answers2

1

Adapted from https://proofwiki.org/wiki/Projection_on_Real_Euclidean_Plane_is_not_Closed_Mapping:

Take $\mathbb R^2$ with the usual topology, and the set $S=\{(x,y)\in\mathbb R^2\mid xy=1\}$. $S$ is obviously closed in $\mathbb R^2$, but its projection on the $x$ axis: $(-\infty,0)\cup(0,\infty)$ is not.

1

"My attempt is to see if I might generate a closed product space whose projection maps to clopen sets."

The problem with this approach is that clopen sets are nevertheless closed sets. So this approach won't work for finding counterexamples of a projection that takes a closed set to a set that is not closed.

"I'm working backwards, that is, from the decompositions to the product space."

This is good intuition. It's exactly how I'd approach this myself.

Finally, $[0,1) \times (0,1]$ is the set $\{(x,y) \in \mathbb{R}^2 \ | \ 0 \leq x < 1 \text{ and } 0 < y \leq 1 \}$. Draw a picture according to these rules; you'll see that it's a square with its right and bottom boundary edges removed. This alone won't work as a counterexample though since $[0,1) \times (0,1]$ isn't itself closed, but maybe you could take it under the subspace topology and construct something inside that works.

One thing you could do: start with your favorite countably infinite not-closed set $V = \{x_k\}_{k=1}^\infty$ in $\mathbb{R}$, then construct the set $U = \Big\{(x_k, k) \Big\}_{k=1}^\infty \subset \mathbb{R}^2$. Notice that the points of $U$ are "far" from each other (by a distance of at least $1$ thanks to the $y$-coordinate)—i.e. every point in $U$ is isolated. From this, it's easy to see that $U$ is closed, but the projection of $U$ onto $\mathbb{R}$ returns $V$, the not-closed set that we started with.

Side remark: Functions that map closed sets to closed sets are called closed maps. There is a theorem that states that a continuous function from a compact space to a Hausdorff space is necessarily closed. Moreover, projection maps on a product space with the product topology are, by definition, continuous. Therefore, if you're interested in finding more counterexamples, make sure that your domain is not compact if your codomain is going to be Hausdorff.

Kaj Hansen
  • 33,011
  • Working through your example has exposed a bunch of hole in my understanding. If I follow the construction correctly, and if I choose $\mathbb{N}$ as a countable subset of $\mathbb{R}$, then the set $U$ will look like points $(1,1), (2,2)...$, which is open in the induced topology from $\mathbb{R}^2$ because the isolated points in $\mathbb{R}^2$ would have plenty of open sets in the product space. Further, am I right in thinking that the images of the projections would simple be the natural numbers, which are closed? And so, we have a projection that maps an open set to a closed set. – innumeratus Feb 18 '18 at 17:25
  • I think there could be a misunderstanding between us about what you're trying to prove (or maybe not just making sure). My discussion above is about finding an example of a product space $\displaystyle \prod_k X_k$ and a projection $\pi_j: \displaystyle \prod_k X_k \rightarrow X_j $ in which a closed set is mapped to a set that's not closed. – Kaj Hansen Feb 19 '18 at 06:39
  • $\mathbb{N}$ is a closed set, whereas I'm saying to start with a set that is not closed. That aside, are you considering $\mathbb{N} \times \mathbb{N}$ as a potential closed set to look at? If so, the subspace topology is not what you want to consider, but rather think of $\mathbb{N} \times \mathbb{N}$ as a subset of $\mathbb{R}^2$. If I'm understanding correctly, I think the goal is to find a closed set relative to some overlying topology--i.e. some $\displaystyle U_k \subsetneq \prod_k X_k$--and then hope that $\pi_j(U_k)$ is not closed in $X_j$. – Kaj Hansen Feb 19 '18 at 06:40
  • Everything you said is true; it just doesn't solve what I'm interpreting the problem to be. If you consider $X = \mathbb{N} \times \mathbb{N}$ and then take the closed set to be $X$ itself, then it's true that $\pi(X) = \mathbb{N}$, which is both closed and open because its the whole space. That latter fact is true about $X$ as well. So you've given an example that takes a closed set to an closed set. – Kaj Hansen Feb 19 '18 at 06:49
  • Sorry for being so verbose. I should've addressed this in my original post, but you say: "My attempt is to see if I might generate a closed product space whose projection maps to clopen sets". The problem with this approach is that clopen sets are nevertheless closed sets. So this approach won't work for finding counterexamples of a projection that takes a closed set to a set that is not closed. When you say: "I'm working backwards, that is, from the decompositions to the product space". This is good intuition; it's exactly what I've done in my approach. – Kaj Hansen Feb 19 '18 at 06:52