0
$\begingroup$

Define $\lim\limits_{\leftarrow n}X_n\ $to be the subset of $\prod_{n\geq 1}^{}X_n$ constisting of elements $(x_1,x_2,...)=(f_1(x_2),f_2(x_3),...)$.

Show that if each $X_n$ is equipped with the discrete topology and $\prod_{n\geq 1}^{}X_n$ with the corresponding product topology, then $\lim\limits_{\leftarrow n}X_n\ $ is Hausdorff.

So far for two distinct points $x=(x_1,x_2,...)=(f_1(x_{2}),f_2(x_{3}),...)$ and $y=(y_1,y_2,...)=(f_1(y_{2}),f_2(y_{3}),...)$, I have constructed the open sets $$U=\{f_1(x_2)\}\times X_2\times X_3..,\,\,\,\,\, V=X_1\times \{f_2(y_3)\}\times X_3,...$$ but I have trouble showing that they are disjoint, hence $\lim\limits_{\leftarrow n} X_n\cap U\cap \lim\limits_{\leftarrow n} X_n\cap V$ disjoint in the subspace topology.

This is the right step right?

  • 2
    Can you state clearly what you are trying to prove? – 2017-01-15
  • 0
    Consider a countable collection of sets ${X_n}\geq 1$ and functions $f_n:X_{n+1}\to X_n$ We could organize these into a diagram that looks as follows: · · · $X_3\to X_2\to X_1$ Define $\lim\limits_{\leftarrow n} X_n$ (the ‘inverse limit’ of the Xn’s) to be the subset of $\prod_{n\geq 1}^{}X_n$ consisting of those elements (x1, x2, . . .) for which $x_n$ = $f_n(x_{n+1})$. – 2017-01-15
  • 3
    I would suggest proving more generally that an arbitrary product of Hausdorff spaces is Hausdorff and that any subspace of a Hausdorff space is Hausdorff. (The topology on an inverse limit is the supspace topology of the product toplogy.) – 2017-01-15
  • 0
    @PedroTamaroff Sorry, it should be better now. – 2017-01-15

0 Answers 0