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Let $\{0,1\}$ be equipped with the Sierpiński topology $\{\emptyset, \{0,1\},\{1\}\}$, and $\mathbb{R}^d$ with the usual Euclidean topology. Then is the pointwise-convergence (point-open) topology on $C(\mathbb{R}^d,\{0,1\})$ indeed weaker than the compact-open topology?

I have in mind the case where $n>1$.

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    $\begingroup$ It's "Sierpinski" (Sierpiński). Actually "Sherpinski" is probably a better transliteration for pronouncing, but is seldom or never used (for him). But definitely it's not "-isnki". Also "product", not "produce". $\endgroup$
    – YCor
    Apr 17, 2020 at 11:19
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    $\begingroup$ @YCor I don't know enough English grammar, but I think the correct way is to write it exactly as it is written in Polish. I believe transliteration can be only used for languages with non-latin alphabets. $\endgroup$ Apr 17, 2020 at 11:38
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    $\begingroup$ @მამუკაჯიბლაძე But "Sherpinski" does exist (not for the mathematician), so it has most likely been transliterated at some point. Many surnames of foreign origin have been transliterated at some point: one can prefer to change the spelling than the pronunciation. Even in our case, one can see "Sierpinski" as a widespread transliteration of "Sierpiński" (to write the latter I have to copy-paste from a text including the 'ń' character, so don't always do it). $\endgroup$
    – YCor
    Apr 17, 2020 at 11:54
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    $\begingroup$ Doesn't the question reduce to the case $n=1$? A function into a product is continuous if and only if so are the components (=compositions with the projections). $\endgroup$ Apr 17, 2020 at 15:12
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    $\begingroup$ @EmilJeřábek Hard to say. Russian pronunciation of the name Sierpiński has S rather than Sh, it sounds like Serpinskij (Серпинский). Another example - Łoś is properly pronounced like "wash", but with "sh" softened like the first sound in Sierpiński (ɕ). Russian pronunciation is Los' (Лось), with softened s. The word means moose both in Polish and in Russian. But I think you agree that quite probably English pronunciation stems from transliteration rather than from Russian version of the word. $\endgroup$ Apr 18, 2020 at 18:33

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Functions $\mathbb R^d\to \{0,1\}$ are indicator functions $f=I_{A}$ with $A=f^{-1}(\{1\})$ and continuity with respect to the Sierpiński topology precisely means that $A$ is open in $\mathbb R^d$. The topology of pointwise convergence on $C(\mathbb R^d,\{0,1\})$ is strictly coarser than the compact-open topology: Since the only interesting open set in $\{0,1\}$ is $\{1\}$ the sets $W(K)=\{I_B: K\subseteq B\}$ ($K$ compact and $B$ open in $\mathbb{R}^d$) form a base of the compact-open topology and the sets $W(E)$ with finite $E$ form base of the point-open topology (that name is probably not very common).

To show that the claim one has to find a compact set $K$ such that $W(K)$ does not contain any $W(E)$ for a finite set $E$. This is the case, e.g, for the closed unit ball $K$ in $\mathbb R^d$: For any finite $E$ take a union $B$ of smalls open balls centered at the points of $E$ so that the volume of the $B$ is strictly less than the volume of $K$. Then $I_B \in W(E) \setminus W(K)$.

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  • $\begingroup$ What does it exactly mean for a sequence of indicators $I_{B_n}$ to converge to an indicator ($B_n,B$ open subsets of $\mathbb{R}^d$) in $C(\mathbb{R}^d,\{0,1\})$? I'm having trouble interpreting. $\endgroup$
    – ABIM
    Feb 1, 2021 at 20:08

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