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This is an exercise from Mendelson's Introduction to Topology, page 101.

THEOREM

For each $n\in \Bbb N$, let $X_n$ be the discrete two-point topological space $\{0,2\}$. Define the product space $X=\prod\limits_{n\in \mathbb N}X_n\;$. Let $f:X\to [0,1]$ be defined as

$$f(x)=\sum_{n=1}^\infty \frac{x(n)}{3^n}$$

Then $f$ is one-one and continuous.

PROOF (Revised)

We first prove $f$ is one-one. To this end, suppose we had $$\sum_{k\geqslant 1}\frac{x_n}{3^n}=0$$ with $x_n\in\{0,-2,2\}$. I claim first there cannot be any $K$ with $x_K=-2$. Indeed, let $K$ be the least index with $x_K=-2$. I claim that $x_k=0$ if $k<K$. Indeed, because $$ - \sum\limits_{k = K}^\infty {\frac{2}{{{3^k}}}} = - \frac{1}{{{3^K}}}\sum\limits_{k = K}^\infty {\frac{2}{{{3^{k - K}}}}} = - \frac{2}{{{3^K}}}\sum\limits_{k = 0}^\infty {\frac{1}{{{3^k}}}} = - \frac{1}{{{3^K}}}$$

there is no way the sum would equal zero if some $x_k$ with $k<K$ was nonzero. But with this out of the way, it is impossible the sum is zero. By the same token, in the extreme case we have $$ - \frac{2}{{{3^K}}} + \sum\limits_{k = K + 1}^\infty {\frac{2}{{{3^k}}}} = - \frac{2}{{{3^K}}} + \frac{1}{{{3^{K + 1}}}} < 0$$

This means there cannot exist $k$ with $x_K=-2$. Thus $x_k\in \{0,2\}$. Since the sum is zero, all nonnegative terms must be zero, so the sequence is identically zero.

We now prove $f$ is continuous. Let $a\in X$ and $\epsilon >0$ be given. The claim is that given the open ball $B(f(a);\epsilon)$, we can choose $k$ so that

$$\tag 1 P_k=\bigcap_{i=1}^k p_i^{-1}(a(i))\subset f^{-1}(B)$$

where $p_i:X\to \{0,2\}$ is the projection to the $i$th coordinate. Since $f$ is one-one, $(1)$ is the same as

$$f(P_k)\subset B$$

Note that for any point $x\in P_k$, the difference $|f(x)-f(a)|$ is at most $3^{-k}$, precisely when $x(n)=2$ and $a(n)=0$ for $n\geq k+1$ (or viceversa):

$$\sum\limits_{n = k + 1}^{ + \infty } {\frac{2}{{{3^n}}}} = \frac{1}{{{3^k}}}$$

Thus we may choose $k$ such that $3^{-k}<\epsilon$. It will follow that $f(P_k)\subset B$, and $f$ will be continuous.

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@tomasz Fixed, sorry. –  Pedro Tamaroff Aug 11 '12 at 19:18
    
How do you define the Cantor space, by the way? –  Asaf Karagila Aug 11 '12 at 19:25
    
@AsafKaragila ${}{}f(X){}{}$ is an option. –  Pedro Tamaroff Aug 11 '12 at 19:25
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2 Answers

up vote 3 down vote accepted

Hint: Pick $N$ so that $$\sum_{n=N}^\infty \frac{1}{3^n}<\epsilon$$

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@PeterTamaroff The hint? or your approach? –  azarel Aug 11 '12 at 22:12
    
@PeterTamaroff A basic open set about $x$ consists in all sequences $y$ that agree with $x$ up the first $n$ elements. So you $|f(x)-f(y)|$ is the tail of the series you mentioned above. –  azarel Aug 11 '12 at 22:22
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This is not true. Consider $x=(0,1,1,0,???\ldots)$. Then for $\varepsilon=1/27$ we have $N=1$, but then if we take $x'=(0,0,0,0,0,\ldots)$, then $f(x)-f(x')\geq 1/3$. In fact, using similar method, we can make $\varepsilon$ arbitrarily small without changing $N$.

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