For a closed subset $F ⊂ X × Y$ , the image $π(F )$ need not be closed in Y

Question

I have the following question from an exam

If $(X, d_X)$ is compact, show that every sequence in $X$ has a subsequence converging to a point of $X$. Deduce that the projection map $\pi$ then has the property that, for any closed subset $F \subset X \times Y$, the image $\pi(F)$ is closed in $Y$. Give an example to show that this fails if $(X, d_X)$ is not assumed compact.

I was able to do the first two parts just fine, however, I am struggling to find a counter example to the last part. I think that it is due to an unease with dealing with the open and closed sets in the product topology. That is, whenever I make up some product topology I find it hard to immidietly tell if a set is open or closed, even if it the product topology as simple components.

Regardless, I tried to construct a counter example. First I thought of some the graph of some curve, however, if the curve is continuous then I will never be able to find an example as desired.

I am getting the sense that this problem will be hard with $\mathbf{R}$, however, I am not sure what other space to try.

Question: How does one think of the last part of the problem and what is such an example?

Answer 1

Let $X=\omega$ be the nonnegative integers with the discrete topology and $Y=\omega\cup\{\infty\}$ be its one point compactification. Then $H=\{(n,n):n\in\omega\}$ is closed in $X\times Y$ but $\pi_Y[H]=\omega$ is not closed in $Y$.

Answer 2

Consider the subsset $F = \{(n,\frac{1}{n}) \mid n \text{ is an integer greater than 1}\}$ of $\Bbb R \times \Bbb R$.

For integer $k \ge 1$ define

$\quad U_k = \{((x,y) \in \Bbb R \times \Bbb R \mid k \lt x \lt k+1\}$

$\quad V_k = \{((x,y) \in \Bbb R \times \Bbb R \mid (k+1)^{-1} \lt y \lt k^{-1}\}$

and define

$\quad U_0 = \{((x,y) \in \Bbb R \times \Bbb R \mid -\infty \lt x \lt 1\}$

$\quad V_0 = \{((x,y) \in \Bbb R \times \Bbb R \mid 1 \lt y \lt +\infty\}$

Clearly for $k \ge 0$ each of the sets $U_k$ and $V_k$ is open. Moreover, it can be shown that these sets along with the set $F$ partitions $\Bbb R \times \Bbb R$. It follows that $F$ is closed in $\Bbb R \times \Bbb R$, but the projection $\pi_y$ onto the $2^{nd}$ coordinate sends $F$ to

$$ \{k^{-1} \mid k \ge 1\}$$

which is not a closed set in $\Bbb R$.

Answer 3

Let me also add one example that I like. Consider $X=\mathbb{R}=Y$ and

$$ F=\{ (x,e^x) \in \mathbb{R}^2 \ : \ x\in \mathbb{R} \}. $$

This is closed. Indeed, pick a sequence $(x_n,y_n)_{n\geq 1} \subseteq F$ such that $\lim_{n\rightarrow \infty} (x_n,y_n)=(u,v)\in \mathbb{R}^2$. By definition this means $\lim_{n\rightarrow \infty} x_n =u, \lim_{n\rightarrow \infty} y_n =v$. However, as $(x_n,y_n)\in F$ we know that $y_n = e^{x_n}$ and hence $$ v=\lim_{n\rightarrow \infty} y_n = \lim_{n\rightarrow \infty} e^{x_n} = e^u $$ as the exponential is continuous. Hence, we get that $(u,v)=(u,e^u)\in F$ and therefore $F$ is closed.

Now let's project $F$ on the second coordinate. We get

$$ \pi(F) = \{ e^x \ : \ x\in \mathbb{R}\} = (0,\infty) $$

which is not closed.