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I'm new to topology, and can't figure out why the metric and product topologies over $\mathbb{R}^n$ are equivalent. Could someone please show me how to prove this?

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2 Answers 2

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The product topology is induced by this norm. $$\|x\|_{\rm prod} = \max\{|x_k|, 1\le k \le n\}$$ Let us use $\|\cdot\|$ for the Euclidean norm. Then $$\|x\| = \left(\sum_{k=1}^n x_k^2\right)^{1/2}\le \left(\sum_{k=1}^n \|x\|_{\rm prod}^2\right)^{1/2} = \|x\|_{\rm prod}\sqrt{n}.$$

Now for a reverse inequality.
We have $$|x_k |\le \left(\sum_{k=1}^n x_k^2\right)^{1/2}, \qquad 1\le k \le n,$$ so $$\|x\|_{\rm prod} \le \|x\|.$$

The norms are equivalent.

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Thanks, Byron. Repair has been instituted. –  ncmathsadist Jun 24 '12 at 19:31
    
Thank you. This was very helpful. –  Parakee Jun 24 '12 at 19:32

In order to prove that two topologies are equivalent you need to prove that for each point $x$ and each element open set of the first topology $U_1$ such that $x\in U_1$ there exist some some open set of the second topology $U_2$ such that $x\in U_2\subset U_1$ and vice versa.

Take $x\in\mathbb{R}^n$ and take some open set $U_1$ of metric toplogy that contains $x$. Since $U_1$ is open we can find a ball $B(x,r)\subset U_1$ that conatins $x$. This ball contains the box $U_2=[x-\delta,x+\delta]\times\ldots\times[x-\delta,x+\delta]$ for $\delta$ small enough (in fact we can take $\delta=n^{-1/2}r$). which is an open set in the product topology and moreover $x\in U_2\subset B(x,r)\subset U_1$. Hence we proved that metric topology is contained in the product topology.

Again take $x\in\mathbb{R}^n$ and take some open set $U_2$ of the product toplogy that contains $x$. Since $U_2$ is open we can find a box $[x-\delta,x+\delta]\times\ldots\times[x-\delta,x+\delta]$ which contains $x$. This box contains the ball $U_1:=B(x,0.5\cdot\delta)$ i.e. an element of the base of the metric topology and moreover $x\in U_1\subset[x-\delta,x+\delta]^n\subset U_2$. Thus we proved that product topology is contained in the metric topology.

Since both implications are proved we conclude that this topologies are coincide.

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Not exactly. You need to show that for every $x$ and every set open $U$ in the first topology containing $x$, there is an open set $V$ in the second topology which is not only a subset of $U$, but also contains $x$ –  tomasz Jun 24 '12 at 19:19
    
Thanks for your attention, I will edit it –  Norbert Jun 24 '12 at 19:20
    
For a concrete example where it is necessary: take $[0,1]\cup \lbrace 0'\rbrace$ with two topologies which agree when restricted to $[0,1]$ or $\lbrace 0'\rbrace\cup (0,1]$ (both restrictions homeomorphic to $[0,1]$), but where one distinguishes $0$ and $0'$ (that is, $[0,1]$ and $(0,1]\cup\lbrace 0'\rbrace$ are both open), and the other doesn't (that is, for any open set $U$, $0\in U \iff 0'\in U$). –  tomasz Jun 24 '12 at 19:28
    
That's how I would prove it,Norbert. But the "analytic" proof above given by ncmathsadist is equivalent because proving that the 2 norms are equivalent proves that each open set containing the point x in the base generated by each norm contains an open subset also containing x generated by the other norm.Which of course is a special case of exactly what your proof says. –  Mathemagician1234 Jun 24 '12 at 20:08
    
Well honestly that proof is not topological but the OP gave a wink that he expects a topological approach. If not this restriction the proof may be much more simple - identity between $\mathbb{R}^n$ with this two topologies is a homeomorphism. –  Norbert Jun 24 '12 at 20:11

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