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I have been told that (in context of Tychonoff's theorem) that $\prod_{i\in I}X_i$ (take for example $I=\{1,2\}$) and $X_1 \times X_2$ are isomorphic. Generally when $I$ is countably infinite or finite then $\prod_{i\in I}X_i$ is isomorphic to $\prod_{i=1}^\infty X_i$.

I am not able to understand how they are isomorphic. My main issue is not being able to understand the what "product space" means and how is it different from the product topology. Waiting for a reply. Thanks

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The notion $\Pi$ is just to shorten the product symboling. If I understood correctly what you mean, then they are not only isomorphic (did you mean homeomorphic?) but identical. I.e. $\Pi_{i\in I}X_{i}=\Pi_{i=1}^{2}X_{i}=X_{1}\times X_{2}$ (where $I=\{1,2\}$). Just a matter of notation. –  Thomas E. May 17 '12 at 13:19
    
It seems that this question is more about understanding what the Cartesian product $\prod_{i\in I} X_i$ (of sets) means, that's why I added elementary-set-theory tag. –  Martin Sleziak May 17 '12 at 13:48
    
They both stand for the same thing: the Cartesian product of the underlying point sets of the $X_i$, given the product topology formed from the topologies of the $X_i$ factors. –  anon May 17 '12 at 14:02
    
I am particularly referring to the link topologywithouttears.net/topbook.pdf . Page number 224. can anyone check and see if at all there is some difference . Thanks. –  Theorem May 17 '12 at 14:18
    
A product space is a space with the product topology. –  Joe Johnson 126 May 17 '12 at 14:25
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1 Answer

up vote 2 down vote accepted

Going from $n$-tuples $(x_1,\cdots,x_n)$ or infinite tuples $(x_1,x_2,\cdots)$ to restricted functions $I\to \bigcup_{i\in I}X_i$, for $I$ an arbitrary index set, is a fundamental change in notation. Notice that with the usual tuples, we can envisage them as these very functions. In particular, given $x=(x_1,x_2\cdots)$, we then have

$$f_x:\{1,2,\cdots\}\to \bigcup_{i=1} X_i: ~ i\mapsto x_i.$$

In other words, any tuple has associated to it a function from the index set to the union of point sets which sends an index $i$ to the component of the tuple indexed by $i$, ie $x_i$. Conversely, any tuple can be easily recovered from its associated function. The reason these maps make such great avatars for the tuples is that, unlike tuples, they do not require the index set to be countable, nor do they require the index set have any linear ordering (which is what allows us to think of tuples as having a first coordinate, and then a second, and a third, etc.). They generalize tuples in this way.

So, for the sake of the argument, let $I=\{1,2\}$ with topological spaces $(X_1,\tau_1)$ and $(X_2,\tau_2)$. Plus, define $\prod_{I}X_i$ to contain functions $f:\{1,2\}\to X_1\cup X_2$ such that $f(1)\in X_1$ and $f(2)\in X_2$. On top of this, the latter point set has a topology: for each pair $O_1\in\tau_1,O_2\in\tau_2$, the set of all functions such that $f(1)\in O_1$ and $f(2)\in O_2$ is an open set of $\prod_IX_i$.

Then there is an "isomorphism" of topological spaces induced by the map on the point sets, namely given by $x\mapsto f_x$. The inverse of the map is simple: send the function $f$ to the tuple $(f(1),f(2))$. We should be clear on what we mean by an "isomorphism," though. Suppose $(A,\varphi)$ and $(B,\psi)$ are two topological spaces. We will say $\rho:A\to B$ induces an isomorphism if it is a bijection of sets, and furthermore $\rho(U)\in \psi$ and $\rho^{-1}(V)\in\varphi$ for all open sets $U\in\varphi,V\in\psi$. That is, the open sets of $B$ are precisely the images of open sets of $A$ under $\rho$, and conversely the open sets of $A$ are the images of the open sets of $B$ under the inverse $\rho^{-1}$ (or, equivalently, the preimages of $B$'s open sets under $\rho$).

The reason we have an "isomorphism" is because

  1. The map $x\mapsto f_x$ is bijective, as it has a well-defined inverse, $f\mapsto(f(1),f(2))$.
  2. If $x_1\in O_1\in\tau_1$ and $x_2\in O_2\in\tau_2$, then $f_x$ is contained in the open set of functions sending $1\in I$ to somewhere in $O_1$ and $2\in I$ to somewhere in $O_2$. And vice versa. This establishes a correspondence between the points of $O_1\times O_2$, an open set in the first topology, and the elements of the open set containing functions sending $1$ into $O_1$ and $2$ to $O_2$. Therefore the images of the open sets in the first are open sets in the second topology and vice-versa.

The same thinking works with $I=\Bbb N$ too.

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thanks a lot , very nice. –  Theorem May 17 '12 at 15:41
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