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Is there any theorem (with reasonably weak conditions) that guarantees that the topology defined by one metric is not finer/coarser than the topology defined by the other?

I am reading Remmert's Theory of Complex Functions. On p.22, exercise 1, Remmert asks whether on the set of all bounded sequences in $\mathbb C$, the open sets defined by the two metrics $d_1(\mathbf a,\mathbf b)=\|\mathbf a-\mathbf b\|_\infty$ and $d_2(\mathbf a,\mathbf b)=\sum_{k=0}^{\infty}2^{-k}|a_k-b_k|$ coincide.

Since $d_2\le 2d_1$, we immediately see that open sets under $d_2$ are open under $d_1$.

The converse is not true, however. It suffices to show that the open unit ball under $d_1$ is not open under $d_2$. Let $\mathbf x$ be the constant sequence $x_k=\frac12$. Clearly it resides in the open unit ball under $d_1$. For any $\epsilon>0$, pick an $n$ such that $2^{-(n+1)}<\epsilon$. Define $y_k=x_k$ when $k\ne n$ and $y_n=1$. Then $d_2(\mathbf x,\mathbf y)=\sum_{k=0}^{\infty}2^{-k}|x_k-y_k|=2^{-n}|\frac12-1|=2^{-(n+1)}<\epsilon$ and $d_1(\mathbf y,\mathbf 0)=1$. In other words, no matter how small $\epsilon$ is, every open ball under $d_2$ with center $\mathbf x$ and radius $\epsilon$ contains some $\mathbf y$ outside the open unit ball under $d_1$. Hence the open unit ball under $d_1$ is not open under $d_2$.

But it took me quite a few minutes to cook up the above counterexample. I wonder if there is some theorem lets me immediately conclude.

Ramen Nii-chan
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  • Actually you proved the desired theorem. But you cannot expexct that there is a general theorem which applies to any twe metrics and lets you immediately conclude something. – Paul Frost Oct 25 '20 at 23:40

3 Answers3

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It's a perfectly nice counterexample. The statement that $d_1$ is not finer than $d_2$ boils down to an existence statement, as you rightly noted, and that is what you proved.

What you did is a perfectly standard method for proving an existence statement, namely: Write down an example and prove that it satisfies the required properties.

It's true, not all existence statements are proved that way. For instance, sometimes one instead quotes an existence axiom to prove an existence statement. This is what happens whenever one applies the Completeness Axiom for the real numbers, for example when proving the Intermediate Value Theorem. Still, though, constructions of new mathematical objects to prove existence statements is an honest living.

Lee Mosher
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Finite dimensional metric spaces are "topologically equivalent" (that is, they are homeomorphic to each other (that is any open ball in one is properly contained in an open ball of the other ) ) because of the Lipschitz inequality between norms that is $c_1|x|_p \leq |x|_q \leq |x|_p \leq c_2 |x|_q$ for any two norms $p \leq q$, where $c_1$ and $c_2$ are well defined constants.

References:

firdaus
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    The set of bounded sequences in $\mathbb C$ is not an example of a finite dimensional metric space, or vector space, or any other kind of finite dimensional space. – Lee Mosher Oct 25 '20 at 22:50
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If the metrics are induced by some norms, I think I have found a sufficient condition (although I cannot say that this condition is reasonably weak).

Proposition. Suppose $\|\cdot\|_1$ and $\|\cdot\|_2$ are two norms on a linear space $X$. If $$\left\{ \|\mathbf x\|_1: \mathbf x\in X,\ \|\mathbf x\|_2=1 \right\}$$ is unbounded above, the open unit ball under $\|\cdot\|_1$ is not open under $\|\cdot\|_2$.

Proof. By the given condition, there exists a sequence of points $\mathbf z_n\in X$ such that $\|\mathbf z_n\|_2=1$ and $\lim_{n\to\infty}\|\mathbf z_n\|_1=\infty$. For any $\mathbf x$ in the open unit ball $B_1(\mathbf 0;\|\cdot\|_1)$ under $\|\cdot\|_1$, let $r=\|\mathbf x\|_1\,(<1)$. However, for any $\epsilon>0$, let $\mathbf y_n = \mathbf x + \frac{\epsilon}{2}\mathbf z_n$. Then $\|\mathbf x-\mathbf y_n\|_2=\frac{\epsilon}{2}<\epsilon$, that is, $\mathbf y_n\in B_\epsilon(\mathbf x;\|\cdot\|_2)$ for every $n$, but $$\|\mathbf y_n\|_1 \ge \big| \|\mathbf x\|_1 - \|\mathbf x-\mathbf y_n\|_1 \big| = \big|\,r-\epsilon \|\mathbf z_n\|_1\,\big| \to \infty$$ as $n\to\infty$. In other words, no matter how small $\epsilon$ is, there always exists some $\mathbf y\in B_\epsilon(\mathbf x;\|\cdot\|_2)$ that lies outside $B_1(\mathbf 0;\|\cdot\|_2)$. Therefore $B_1(\mathbf 0;\|\cdot\|_2)$ is not open under $\|\cdot\|_2$.

Application to Remmert's exercise. Clearly the $d_1$ and $d_2$ concerned are induced by the two norms $\|\mathbf x\|_1:=\sup_{n}|x_n|$ and $\|\mathbf x\|_2:=\sum_n2^{-n}|x_n|$. Let $\mathbf z_n$ be the bounded sequence $(0,\ldots,0,2^n,0,\ldots)$ where the power of $2$ appears at the $n$-th position. Then $\|\mathbf z_n\|_2=1$ for all $n$ but $\|\mathbf z_n\|_1=2^n$ is unbounded. Therefore, by the proposition above, the open unit ball under $d_1$ is not open under $d_2$.

Ramen Nii-chan
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