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?
Showing that the product and metric topology on $\mathbb{R}^n$ are equivalent
2 Answers
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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0Thanks, Byron. Repair has been instituted. – 2012-06-24
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0Thank you. This was very helpful. – 2012-06-24
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 topology 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 open 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 topology 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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0Not 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 $x$ – 2012-06-24
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0Thanks for your attention, I will edit it – 2012-06-24
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0For 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 $U$, $0\in U \iff 0'\in U$). – 2012-06-24
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0That'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. – 2012-06-24
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0Well 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. – 2012-06-24