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Let $X$ and $Y$ be topological spaces where $X$ is compact. Show that projection on the second factor $X \times Y \rightarrow Y$ is a closed map.

Give an example where (in the situation above) the projection to the first factor is not a closed map.


Let $p : X \times Y \rightarrow Y$ be the projection and $C \subset X \times Y $ be closed.

Suppose $y \not \in p(C)$.

Now $X \times \{y\} \cap C = \emptyset$, so $X \times Y − C$ is an open set containing $X \times \{y\}$, homeomorphic to $X$ and therefore compact.

For each $x \in X$ choose open sets $U_x \times V_x$ containing $(x, y)$ and contained in $X \times Y - C$.

The sets $U_x, \: x \in X$ forms an open cover of the compact $X$ and so has a finite subcover $U_{x_1}, U_{x_2}, ..., U_{x_n}$.

If $V = \cap_{i=1}^n V_{x_i}$ then $V$ is open and contains $y$.

$U_{x_i} \times V \subset U_{x_i} \times V_{x_i} \subset X \times Y − C \Rightarrow X \times V$ disjoint from $C$.

Now $V$ is disjoint from $p(C)$ and contains $y$.

We conclude $Y − p(C)$ is open and $p(C)$ is closed.

I am having trouble coming up with an example where the projection to the first factor is not a closed map. I would appreciate if anyone could point me in the right direction.

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    $\begingroup$ First, pick a basic case where $X$ is compact and $Y$ is not. Might as well try $X=[0,1]$ and $Y=\mathbb R$, for example, but you could also choose more easily $Y=\mathbb Z^+$. Then find a closed subset $C$ of $X\times Y$ such that $p_1( C)$ is not closed. $\endgroup$ – Thomas Andrews Mar 2 '16 at 0:17
  • $\begingroup$ Thanks @ThomasAndrews for your comment. $\endgroup$ – JKnecht Mar 2 '16 at 2:38
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The projection $(x, y) \mapsto x$ of the plane $\Bbb{R}^2$ onto the real line $\Bbb{R}$ is a useful example of a continuous map that is not closed: subsets of $\Bbb{R}^2$ like the graph of the reciprocal function: $\{(x, 1/x) \mathrel{|} x \neq 0\}$ are closed, but project onto subsets of $\Bbb{R}$ that are not closed. Restricting to a compact set of $x$ values, you find that $G = \{(x, 1/x) \mathrel{|} x \in [-1, 0) \cup (0, 1]\}$ is a closed subset of $[-1, 1] \times \Bbb{R}$ that projects onto an open but not closed subset of the compact set $[-1, 1]$.

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  • $\begingroup$ Thanks Rob for your answer. $\endgroup$ – JKnecht Mar 2 '16 at 2:38
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A slightly simpler answer, though related.

With $X=[0,1]$ and $Y=\mathbb Z^+$, let $$C=\left\{\left(\frac1n,n\right)\mid n\in\mathbb Z^+\right\}$$

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