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I have the following question:

Let $X_i = (\{0,1\}, \operatorname{discrete})$ where $I= 2^{\mathbb{R}}$ the set of index and let $$X= \{0,1\}^{2^{\mathbb{R}}} = \prod_{i \in I}X_i$$

let $D$ a dense set such that $D \subseteq X$ and let the set $$D_i := D \cap p_i^{-1}(1)$$. Prove that for each $i \neq j$ we have that $D_i \neq D_j$

I have tried the following:

Suppose that is true that $D_i = D_j$, then we have that $D \cap p_i^{-1}(1) = D \cap p_j^{-1}(1)$, so inmediatly we have that $ p_i^{-1}(1)= p_j^{-1}(1)$ but this is not true since the projection just will coincide in one point at most.

am I in the correct way?, I am really confused because I don´t know how can I express this more formally

I really appreciate your help

usermath
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user2237939
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3 Answers3

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Let $i \neq j$, both in $I$. Then define a (basic) open set $U$ of $X$ by $U= p_i^{-1}[\{0\}] \cap p_j^{-1}[\{1\}]$ (note that we use that $\{0,1\}$ is discrete here!). Then there is some $d \in D \cap U$, as $D$ is dense so intersects all non-empty open subsets of $X$. Then $d_i = 0, d_j = 1$, so $d \in D_j \setminus D_i$, which shows that $D_j$ and $D_i$ are not equal as sets.

Henno Brandsma
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Let me offer a completely different strategy, one which shows the beauty of non-constructive proofs. This is a proof in three steps, none of which is particularly difficult.

  1. The product of Hausdorff spaces is Hausdorff. Given two points in the product, pick one coordinates where they differ, and construct basic open sets separating these points using the sets which separate the projections onto the chosen coordinate.

  2. If $X$ is a separable Hausdorff space, then $|X|\leq 2^{2^{\aleph_0}}$. If $X$ is separable then there is a countable set $D$ whose limit points are everything. Note that the function which maps $x$ to the set $\{A\subseteq D\mid x\in\overline D\}$ is an injective function (due to the Hausdorff-ness of $X$); and its range is included in $\mathcal{P(P(}D))$.

  3. $2^{2^\Bbb R}$ is too big.

Asaf Karagila
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Proof:

Let $k\in I$, $\quad U_{i} = X_k \times ... \times X_{k} \times ... \times A_{i} \times ... \qquad $ where $A_{i}$ is open in $\{0,1\}$ with the discrete topology in the position $i\in I$

Then $$ U_{i} = X_k \times \dots \times \dots \times \dots \times A_i \times \dots \in \mathcal{T}_{\text{product}}$$ y $$\quad U_{j} = X_{k}\times ... \times ... A_{j} \times ... \times X_{k} \times ... \in \mathcal{T}_{\text{product}}$$

Then $ U_i \cap U_j \in \mathcal{T}_{\text{product}} (i\neq j)$

We have that $U_i = X_k \times \dots \times X_k \times \dots \times \{1\} \times \dots $ is in the i- esima position and $\quad U_j = X_k \times ... \times \{0\} \times ... \times X_{k} \times ...$ is in the j-esima position.

Let $V = U_i \cap U_j $ then $ V \cap D \neq \emptyset$ since $\bar{D} = X$ (D is dense on X).

So $\exists (\alpha _{i}\in X)(\alpha _i\in V\cap D)$ this is $ \alpha _{i} = (\, , \dots,0, \dots 1, \dots)$ where $0$ is in the position i-esima and $1$ is in the e $j-$esima position. So we have that $$\alpha _{i}\in P_{i}^{-1}(1) \wedge \alpha _{i}\in D\Rightarrow \alpha _{i}\in D\cap P_{i}^{-1}(1)\Rightarrow \alpha _{i}\in D_{i}$$

But $\alpha _{i}\notin P_{j}^{-1}(1)$ because in the position $<<j>>$ $\alpha _{i}$ is zero. Then $$\alpha _{i}\notin D_{j}\Rightarrow \forall (i,j \in I )(D_i \neq D_j) \Rightarrow \forall (i\in I)(\alpha _{i}\in D_{i}\subseteq D)\Rightarrow \forall(i\in I)(\alpha _{i}\in D)$$

Since $\forall (i,j \in I)(D_i \neq D_j)$ then for each $D_i$ there is an element in $D$. So D has as much elements like I and this is a contradiction since $D$ is an enumerable set.

The credit for this proof is due to Michael Olmos.

LFRC
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