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I have $X$ as a countable, Tychonoff space, and I want to show that the collection of clopen subset of $X$ form a base for the topology on $X$.

Can I first just define a base $\mathscr{B}$ of X, let $x\in X$ and $E \subset X$ is closed such that $x\notin E$. So $x\in X-E$, which is open.

I also noticed that the interval $[0,1]$ is uncountable. Can I define a function $f: X\to [0,1]$, then f is onto. Do I need to show that $f$ continuous next?

user642796
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Akaichan
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1 Answers1

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Since $X$ is countable, there are no surjective functions $f : X \to [0,1]$. In fact, this is of central importance to the proof (at least as I see it). Perhaps looking at this hint/question will direct you in the right path.

Hint: If $f : X \to [0,1]$ is continuous and $r \in [0,1] \setminus f[X]$, what sort of set is $f^{-1} [\,[0,r]\,]$? $f^{-1} [\,[0,r)\,]$? How do these two sets compare?

Also, note that it suffices to show the following:

Given any $x \in X$ and any open neighbourhood $U$ of $x$ there is a clopen set $V \subseteq U$ which contains $x$.

user642796
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  • so we can just assume that $f$ is continuous? $f^-1[0,r]$ is closed since f is cont. What two sets that I should compare, aren't they the same thing? – Akaichan May 02 '13 at 04:25
  • @IvordesGreenleaf: Well, in my hint $f$ is by assumption continuous. And, yes, those two sets should be the same. The outline of the proof should be as follows: Given $x \in X$ and an open neighbourhood $U$ of $x$ "find" (using complete regularity) a continuous function $f : X \to [0,1]$ so that for an appropriate $r$ we have $0 \in f^{-1} [,[0,r],] \subseteq U$. – user642796 May 02 '13 at 04:42
  • @Arthur : I think you meant $x \in f^{-1}[[0,r]] \subseteq U$. P.s. : It is totally non-standard to use brackets instead of parenthesis to denote maps of functions, i.e. the standard would be $f^{-1}(A)$ instead of $f^{-1}[A]$, or $f(A)$ instead of $f[A]$. – Patrick Da Silva May 02 '13 at 05:13
  • @PatrickDaSilva: Yes, your first observation is correct. As for the second, in the fields that I work in it is quite common to use $f^{-1} [ \cdot ]$ to denote inverse images of sets. For the simple reason that it is not uncommon for elements of a set to also be subsets of the set in question (especially when working with ordinals). So to avoid possible confusion differing notation is used. Of course, if you insist, perhaps you should inform Brian M. Scott that he also uses incorrect notation. – user642796 May 02 '13 at 05:20
  • You mean that you wish to distinguish $f^{-1}(a)$ when $f$ is injective and $f^{-1}$ is the inverse map, and $f^{-1}(A)$ when $A$ is a subset of the codomain of $f$ (so that we would write $f^{-1}(a)$ and $f^{-1}[A]$ or $f(a)$ and $f[A]$)? (Maybe that your field of study uses non-standard notation for the right reasons, but I don't know you so I can't guess that when I read your answer. :P) – Patrick Da Silva May 02 '13 at 05:40
  • Hi, Arthur. Your answer is wonderful. – Paul May 02 '13 at 07:18
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    @Patrick: You may not have seen the notation before, but there is nothing non-standard about using $f^{-1}[A]$ for ${f^{-1}(a):a\in A}$ and $f^{-1}(A)$ for the unique $B\in\operatorname{dom},f$ such that $f(B)=A$ when $A$ is actually an element of the range of $f$. – Brian M. Scott May 02 '13 at 11:36
  • @Arthur Fischer: How do I show that there is a clopen set $V\subset U$ which contains x? And I still can't get it yet, do I need to work with the close set E? – Akaichan May 02 '13 at 13:36
  • @IvordesGreenleaf: This is the heart of the problem, and is what my hint is attempting to direct you towards. Suppose that $U$ is an open neighbourhood of $x$. Can you think of a way to use complete regularity to get a continuous function $f : X \to [0,1]$ such that $x \in f^{-1} [,[0,r),] \subseteq U$ for all $0 < r \leq 1$? – user642796 May 02 '13 at 13:42
  • So you meant to say $f^-1[(o,r)]$ al along, and this set is clopen in X, right? – Akaichan May 02 '13 at 13:47
  • @ArthurFischer:
    And by completely regularity, I have $f(x)\in f(U)=0$ and $f(X−U)=1$, how would that help me though?     – Akaichan May 02 '13 at 14:32
  • @IvordesGreenleaf: Recall that a topological space $X$ is completely normal if given any $x\in X$ and any closed $F\subseteq X$ which does not contain $x$ there is a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $f(z)=1$ for all $z\in F$. What would be really nice is if given $x \in X$ and an open neighbourhood $U$ of $x$, for an appropriately chosen continuous function $f$ there was an $r \in [0,1]$ such that $f^{-1} [,[0,r),]$ where a clopen set containing $x$ and a subset of $U$. We know that $f^{-1}[,[0,r),]$ is always open (by continuity), why would it be also closed? – user642796 May 02 '13 at 17:07
  • @ArthurFischer: because f(x)=0? I don't really get it. Can we just fet that $f^-1[0,r)$ is clopen since $[0,r)$ is clopen? But my question is how to show that x is in $f^-1[0,r)$ and $f^-1[0,r)$ is a subset of U – Akaichan May 02 '13 at 17:33
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    @IvordesGreenleaf: $[0,r)$ is not a clopen subset of $[0,1]$; it is an open subset of $[0,1]$ (since $[0,r) = (-1,r) \cap [0,1]$ and $(-1,r)$ is open in $\mathbb{R}$). (In fact, the only clopen subsets of $[0,1]$ are $\varnothing$ and $[0,1]$ itself.) If $f(x) = 0$, then clearly $x \in f^{-1} [,[0,r)],] = { z \in X : 0 \leq f(z) < r }$. – user642796 May 02 '13 at 17:37
  • @ArthurFischer: And is it correct to say that $f^-1[0,r)$ is a subset of U since $f(X-U)=1$? – Akaichan May 02 '13 at 17:42
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    @IvordesGreenleaf: Yes! – user642796 May 02 '13 at 18:18
  • @ArthurFischer: Phew!! – Akaichan May 02 '13 at 18:50