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I know that the product of two balls of equal radius in metric spaces is not necessarily a ball in the product space.

enter image description here

But I couldn't identify the fault in the proof where I showed

$B_X(a,\epsilon)\times B_Y(b,\epsilon)=B_{X\times Y}((a,b),\epsilon):$

$(x,y)\in B_X(a,\epsilon)\times B_Y(b,\epsilon)\\\iff x\in B_X(a,\epsilon),y\in B_Y(b,\epsilon)\\\iff d_X(a,x)<\epsilon,d_Y(b,y)<\epsilon\\\iff \max\{d_X(a,x),d_Y(b,y)\}<\epsilon\\\iff d_{X\times Y}((a,b),(x,y))<\epsilon\\\iff(x,y)\in B_{X\times Y}((a,b),\epsilon)$

In the above figure do the topology obtained by defining product metric as $d_{X\times Y}=\max\{d_X,d_Y\}$ different with the topology obtained from $\|.\|_2?$

Sriti Mallick
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  • $\max(d_X,d_Y) < \epsilon \Leftarrow d_{X\times Y} < \epsilon$ only. – user7530 Jul 15 '13 at 03:52
  • Isn't $d_{X\times Y}$ defined as $\max{d_X,d_Y}?$ – Sriti Mallick Jul 15 '13 at 04:00
  • That depends, to define a metric space as a product of $n$ base spaces you need to specify a norm on $\mathbb{R}^n$. Usually this is taken to be the Euclidean norm but you could also specify the max norm, in which case your theorem is true. – user7530 Jul 15 '13 at 04:10
  • So due to the definition I mentioned above my result is true and the answer to the added quoted question is "Yes." Right? – Sriti Mallick Jul 15 '13 at 04:24

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Just because the max distance is less than $\epsilon$ does not mean that the distance in the product space is less than $\epsilon$. Both $x=0.9$ and $y = 0.9$ are within $\epsilon = 1$ of the origin in $\mathbb{R}^1$, but the point $(0.9,0.9)$ is distance $\sqrt{0.9^2 + 0.9^2} = 1.27 \cdots$ from the origin in $\mathbb{R}^2$.

Elchanan Solomon
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