Every totally disconnected space is T1. Is there an example of a space that is totally disconnected but not Hausdorff (T2)? GeoffreyT2000 ( talk, contribs) 02:37, 2 April 2017 (UTC) reply

According to Steen & Seebach, a modified Fort space is just such a space. -- Deacon Vorbis ( talk) 04:12, 2 April 2017 (UTC) reply

Let ${\displaystyle ~\rho (x)~=~{\dfrac {\Gamma (x)\cdot \zeta (2x)}{\pi ^{x}}}~.\quad }$ Then ${\displaystyle ~\rho {\Big (}{\tfrac {1}{4}}+x{\Big )}~=~\rho {\Big (}{\tfrac {1}{4}}-x{\Big )}~}$ constitutes the functional equation for the Riemann ${\displaystyle \zeta }$ function. The presence of the product ${\displaystyle ~\Gamma (x)\cdot \zeta (2x)~}$ is perfectly understandable, inasmuch as the poles of the former coincide with the (trivial) zeroes of the latter; as is also the symmetry with regard to ${\displaystyle {\tfrac {1}{4}},}$ since this value stands midway between ρ's two poles. What poses serious difficulties from an intuitive perspective, however, is the presence of ${\displaystyle ~{\dfrac {\zeta (2x)}{\pi ^{x}}}~}$ instead of the expected ${\displaystyle ~{\dfrac {\zeta (2x)}{\pi ^{\color {red}2x}}}~,~}$ given the fact that ${\displaystyle \zeta (2k)}$ always possesses a known closed form in terms of ${\displaystyle \pi ^{2k}}$ rather than merely ${\displaystyle \pi ^{k}}$ for integer values of the argument k. One can, of course, always write ${\displaystyle ~\rho (x/2)~=~{\dfrac {\Gamma (x/2)~}{~\Gamma (1/2)^{x}}}\cdot \zeta (x),~}$ but, for all its niceness, the latter appears somewhat contrived, inasmuch as the power of ${\displaystyle \pi }$ is clearly a contribution of Riemann's ${\displaystyle \zeta }$ rather than Euler's ${\displaystyle \Gamma }$ function. — 79.113.203.86 ( talk) 14:29, 2 April 2017 (UTC) reply