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April 2

Non-Hausdorff totally disconnected space

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)

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

Intuition Behind the Riemann ζ Functional Equation

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