Relative Coarseness of Topologies from Taxicab and Max Metrics Came across this problem while reading A First Course in Topology: Explaining Continuity by Paul Bankston:

Show that the taxicab metric is strictly finer than the max metric on $C([0,1])$.

Reminder of terminology/notation:


*

*the taxicab norm is: $|f|_t = \int_0^1|f(x)|dx$

*the max norm is: $|f|_m = \max_{x \in [0,1]} |f(x)|$

*the corresponding norm balls are denoted $B_t(x, \epsilon)$ or $B_m(x, \epsilon)$


I managed to convince myself through the steps below of almost exactly the opposite of what the book asks. I'd like to request someone to either point out the flaw in my logic or to confirm that the above question is actually wrong:


*

*I start by using the fact $|f|_t \le |f|_m$ (always).

*Let $\langle X, \mathcal{T}_t\rangle$ = topological space induced by taxicab norm on $X$ and $\langle X, \mathcal{T}_m\rangle$ = topological space induced by max norm on $X$.

*I arbitrarily pick a point $x \in X$ and a $\mathcal{T}_t$-open neighborhood $U$ of $x$:


*

*First pick $\epsilon$ s.t. $B_t(x, \epsilon) \subseteq U$.

*We know that $B_m(x, \epsilon) \subseteq B_t(x, \epsilon)$ because $\forall y \in B_m(x, \epsilon)$:
$$ |x - y|_t \le |x-y|_m < \epsilon \implies y \in B_t(x, \epsilon)$$

*This means each $\mathcal{T}_t$-open set can be represented as a union of $\mathcal{T}_m$-open sets, which is equivalent to saying each $\mathcal{T}_t$-open set is itself a $\mathcal{T}_m$-open set. This means $\mathcal{T}_t \subseteq \mathcal{T}_m$. And thus the taxicab metric is actually coarser than the max metric.



If all this is true, then the taxicab metric cannot be strictly finer than the max metric.
 A: Yes, the book is in error. 

As you concluded, the comparison should be reversed.

Claim: On $C[0,1]$, the max metric is strictly finer than the taxicab metric.

To prove the above claim, it suffices to show two things.


*

*For any $f \in X$, and any open ball centered at $f$ in the taxicab metric, $B_t(f,\epsilon)$ say, there is some open ball centered at $f$ in the max metric, $B_m(f,\delta)$ say, such that $B_m(f,\delta) \subseteq B_t(f,\epsilon)$.$\\[4pt]$

*For some $f \in X$, there is some open ball centered at $f$ in the max metric, $B_m(f,\epsilon)$ say, for which there is no open ball centered at $f$ in the taxicab metric, $B_t(f,\delta)$ say, such that $B_t(f,\delta) \subseteq B_m(f,\epsilon)$.


Start with objective $1$ . . .

Let $f \in X$, and let $\epsilon > 0$. 

Claim $B_m(f,\epsilon) \subseteq B_t(f,\epsilon)$.
\begin{align*}
&g \in B_m(f,\epsilon)\\[4pt]
\implies\;&|g(a) - f(a)| < \epsilon,\;\text{for all}\;a \in [0,1]\\[4pt]
\implies\;&\int_0^1|g(x) - f(x)|dx < \epsilon\\[4pt]
\implies\;&g \in B_t(f,\epsilon)
\end{align*}
so $B_m(f,\epsilon) \subseteq B_t(f,\epsilon)$, as claimed. 

Thus, using $\delta = \epsilon$, objective $1$ is achieved.

Next, the second objective . . .

Let $f=0$, and let $\epsilon=1$.

Let $\delta > 0$, and let $g$ in $X$ be defined by
$$
g(x) =
\begin{cases}
\!\bigl(-{\large{\frac{4}{\delta}}}\bigr)x + 2&\;\;\;\text{if}\;0 \le x \le \min\bigl({\large{\frac{\delta}{2}}},1\bigr)
\qquad\qquad\qquad\;\;
\\[4pt]
\;\;0&\;\;\;\text{otherwise}
\end{cases}
$$
\begin{align*}
\text{Then}&\int_0^1|g(x) - f(x)|dx\\[4pt]
=&\int_0^1g(x)dx&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\text{[since $f=0$, and $g$ is nonnegative]}\\[4pt]
\le&\int_0^\delta-\bigl({\small{\frac{4}{\delta}}}\bigr)x + 2\;dx\\[4pt]
=&\;\,{\small{\frac{\delta}{2}}}\\[4pt]
 < &\;\,\delta\\[4pt]
\end{align*}
Hence $g \in B_t(0,\delta).\;$

But $g(0)=2$, so $g \notin B_m(0,1).\;$

Since $\delta > 0$ was arbitrary, it follows that there does not exist $\delta > 0$ such that $B_t(0,\delta) \subseteq B_m(0,1)$.

Thus, objective $2$ is achieved.

This completes the proof.
