Let $\prod_{\alpha \in J} X_{\alpha}$ is the product of typologies. Consider product topology on the set. Then any function $f : A \rightarrow \prod_{\alpha \in J} X_{\alpha}$ will be continuous on the space $\prod_{\alpha \in J} X_{\alpha}$ if and only if each of its components $f_{\alpha}$ be continuous on the topology $X_{\alpha}$. Can we say one part (if condition) of this result is true for box topology $\prod_{\alpha \in J} X_{\alpha}$ ?

A sequence $\{x^n\} = (x_i^n)_{i \in J, n \in \mathbb{N}}$ in $\prod_{\alpha \in J} X_{\alpha}$ will be convergent in product topology iff each of the component sequences $\{x_i^n\}_{n \in \mathbb{N}}$ in the space $X_i$ is convergent. Is the result true for box topology? What will be difference between the closers of sets in the product and box topology and how to find them?

Please help me a little to understand these basic difference between the box and the product topology.

Thank you.


I will write $\square_{\alpha\in J}X_\alpha$ for the box product and $\prod_{\alpha\in J}X_\alpha$ for the ordinary Tikhonov product. Suppose that $f:A\to\square_{\alpha\in J}X_\alpha$ is continuous. The box topology is finer than the Tikhonov topology, so automatically $f$ is continuous as a function from $A$ to $\prod_{\alpha\in J}X_\alpha$, and therefore each component map $\pi_\alpha\circ f$ is continuous as a function from $A$ to $X_\alpha$.

The question of convergent sequences is a bit more complicated. To avoid trivial technical difficulties, assume that $J$ is infinite, and each $X_\alpha$ is a $T_1$-space with at least two points. Let $X=\square_{\alpha\in J}X_\alpha$. For $x=\langle x_\alpha:\alpha\in J\rangle,y=\langle y_\alpha:\alpha\in J\rangle\in X$ let $$D(x,y)=\{\alpha\in J:x_\alpha\ne y_\alpha\}\;.$$

Suppose that $x=\langle x_\alpha:\alpha\in J\rangle\in X$ and $x^n=\langle x_\alpha^n:\alpha\in J\rangle\in X$ for $n\in\Bbb N$, and that $\langle x^n:n\in\Bbb N\rangle$ converges to $x$ in the box product $X$; I claim that there are an $m\in\Bbb N$ and a finite $F\subseteq J$ such that $D(x^n,x)\subseteq F$ for all $n\ge m$.

Suppose not; then we can choose an $n_0\in\Bbb N$ and an $\alpha_0\in J$ such that $x_{\alpha_0}^{n_0}\ne x_{\alpha_0}$. Suppose that $m\in\Bbb N$, and for $k=0,\dots,m$ we’ve chosen $n_k\in\Bbb N$ and $\alpha_k\in J$ in such a way that $n_0<\ldots<n_m$, $x_{\alpha_k}^{n_k}\ne x_{\alpha_k}$, and the indices $\alpha_0,\dots,\alpha_m$ are all distinct. By hypothesis there is no $\ell\in\Bbb N$ such that $D(x^i,x)\subseteq\{\alpha_0,\dots,\alpha_m\}$ for all $i\ge\ell$, so there are an $n_{m+1}\in\Bbb N$ and an $\alpha_{m+1}\in J\setminus\{\alpha_0,\dots,\alpha_m\}$ such that $n_{m+1}>n_m$ and $x_{\alpha_{m+1}}^{n_{m+1}}\ne x_{\alpha_{m+1}}$. In this way we recursively construct $\{n_k:k\in\Bbb N\}\subseteq\Bbb N$ and $A=\{\alpha_k:k\in\Bbb N\}\subseteq J$ so that $\langle n_k:k\in\Bbb N\rangle$ is strictly increasing, $x_{\alpha_k}^{n_k}\ne x_{\alpha_k}$ for each $k\in\Bbb N$, and if $k,\ell\in\Bbb N$ with $k\ne\ell$, then $\alpha_k\ne\alpha_\ell$.

For each $k\in\Bbb N$ let $U_{\alpha_k}$ be an open nbhd of $x_{\alpha_k}$ that does not contain $x_{\alpha_k}^{n_k}$, and for $\alpha\in J\setminus A$ let $U_\alpha=X_\alpha$. Let $U=\square_{\alpha\in J}U_\alpha$; clearly $U$ is an open nbhd of $x$ in $X$. However, for each $k\in\Bbb N$ we have $x_{\alpha_k}^{n_k}\notin U_{\alpha_k}$ and hence $x^{n_k}\notin U$. Thus, the infinite subsequence $\langle x^{n_k}:k\in\Bbb N\rangle$ of $\langle x^k:k\in\Bbb N\rangle$ lies entirely outside $U$, and $\langle x^k:k\in\Bbb N\rangle$ cannot converge to $x$.

If there are an $m\in\Bbb N$ and a finite $F\subseteq J$ such that $D(x^n,x)\subseteq F$ for all $n\ge m$, then $\langle x^k:k\in\Bbb N\rangle$ converges to $X$ iff $\langle\pi_F(x^k):k\in\Bbb N\rangle$ converges to $\pi_F(x)$, where $\pi_F:X\to\prod_{\alpha\in F}X_\alpha$ is the canonical projection map. (Since $F$ is finite, the box and Tikhonov topologies on $\prod_{\alpha\in F}X_\alpha$ are the same.)

This result can be stated as follows: $\langle x^k:k\in\Bbb N\rangle$ converges to $x$ iff there are an $m\in\Bbb N$ and a finite $F\subseteq J$ such that

  1. $x_\alpha^k=x_\alpha$ for all $k\ge m$ and $\alpha\in J\setminus F$, and
  2. for each $\alpha\in F$, $\langle x_\alpha^k:k\in\Bbb N\rangle$ converges to $x_\alpha$ in $X_\alpha$.
  • $\begingroup$ Thank you Sir for your beautiful proof. It gives me some concepts new to me. This will help me to learn Topology from you. $\endgroup$ – Dutta Jul 21 '13 at 10:39
  • $\begingroup$ @Samprity: You’re very welcome. $\endgroup$ – Brian M. Scott Jul 21 '13 at 10:42

The box topology is finer than the product topology; therefore it is harder for a function into a box-topology space to be continuous than for the "same" function into the corresponding product-topology space. (For functions out of the spaces, it's easier for the box-topology version to be continuous.)

For an example, let $A=\mathbb R$ and $X_\alpha=A$ for all $\alpha$, and define $f\colon A \to \prod A$ to be $f(a) = (a,a,a,\dots)$. Then $f$ is actually not continuous in the box topology, even though each component function clearly is. (Consider the inverse image of the open set $(-1,1) \times (-\frac12,\frac12) \times (-\frac13,\frac13) \times \cdots$.)

Similarly, the sequence $\{x^n\}$ where $x^n = (\frac1n,\frac1n,\frac1n,\dots)$ does not converge to $(0,0,0,\dots)$ in the box topology.


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