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I have a question regarding topology of sets in $\mathbb{R^2}$ and $\mathbb{R^3}$. The question is as follows:

For each $A \subset \mathbb{R^3}$, we define $P(A) \subset \mathbb{R^2}$ by \begin{equation} P(A) := \{(x,y)\ |\ \exists\ z \in \mathbb{R}: (x,y,z) \in A\}.\end{equation}

I have the following statements:

"If A is open (in $\mathbb{R^3}$), then $P(A)$ is open (in $\mathbb{R^2}$)."

"If A is closed (in $\mathbb{R^3}$), then $P(A)$ is closed (in $\mathbb{R^2}$)."

"If A is compact (in $\mathbb{R^3}$), then $P(A)$ is compact (in $\mathbb{R^2}$)."

I think that the first statement is true, but I am in doubt about the second one. I was thinking of a set $A = \{(x,y,z)\ | \ x^2 + y^2 < 1, -1 \leq z \leq 1\}$, but I am now sure whether I am going in the correct direction. Could anyone help me out? Thanks in advance!

2 Answers2

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Have a look at the classical projection function:

$$\pi:\mathbb{R}^3\to\mathbb{R}^2$$ $$\pi(a,b,c)=(a,b)$$

and note that

$$P(A)=\pi(A)$$

where on the right side is the image of $A$ via $\pi$.

  • Now the first statement is equivalent to the statement that the projection is an open map which is true.

  • Second statement is not true. Generally projections are not closed maps. For example take

$$A=\bigg\{\big(0, x, \frac{1}{x}\big)\ \bigg|\ x>0\bigg\}$$

This set is closed, however $\pi(A)=\big\{(0,x)\ \big|\ x>0\big\}$ is not.

freakish
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The general def'n of the Tychonoff-product topology on $S(J)=\prod_{j\in J}S_j$ when each $S_j$ is a top'l space is that it is the weakest topology such that for each $a\in J$ the projection $p_a(\;(x_j)_{j\in J}\;)=x_a$ is continuous.

Any $A\subset S(J)$ is open in this topology iff the following holds: For each $x=(x_j)_{j\in J}\in A$, there exists $A(x)=\prod_{j\in J}B_j \subset A$ such that (I)... $x\in A(x)$, and (II)... each $A_j$ is open in $S_j$ and (III)... $\{j\in J: A_j\ne S_j\}$ is finite.

It follows that if $K\subset J$ and $P_K(\;(x_j)_{j\in J}\;)=(x_k)_{k\in K}$ then $P_K$ is a continuous open mapping from $S(J)$ to $S(K)=\prod_{k\in K}S_k.$ (A function is called an open mapping iff the image of each open set is open.)

In particular, when $J$ is finite then $A$ is open in $S(J)$ iff for each $x=(x_j)_{j\in J}\in A$ there exists $A(x)=\prod_{j\in J}A_j\subset A$ with $x\in A(x)$, where each $A_j$ is open in $S_j.$ And hence if $K\subset J$ then the image $P_K(A)$ satisfies $P_K(x)\in \prod_{k\in K}A_k\subset P_K(A),$ so $P_K(A)$ is open in $S(K)$.

For example, in your Q we have $J=\{1,2,3\}$ and $K=\{1,2\}$ and each $S_j=\mathbb R.$ Let $A$ be open in $S_J=\mathbb R^3.$ For any $(x,y,z)\in A\subset S(J)$ there exist open subsets $A_1,A_2,A_3$ of $\mathbb R$ such that $(x,y,z)\in A_1\times A_2\times A_3\subset A .$ So $(x,y)\in A_1\times A_2\subset P_K(A)$. This holds for every $(x,y,z)\in A$ so $P_K(A)$ is open in $S(K)=\mathbb R^2.$

DanielWainfleet
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