3

So here is proved that two continuous functions $f:X\to Y$ and $g:X\to Y$ from a space $X$ to an hausdorff space $Y$ are equal when they agree in a dense set $D$ of $X$. However it seem to me that the result could hold also when $Y$ is not hausdorff as the following argumentations seem to show.

So if $x\in X\setminus D$ then $x\in\operatorname{cl}D$ and so that it is the limit of a net $(x_\lambda)_{\lambda\in\Lambda}$ in $D$ and thus by continuity we conclude that $$ f(x)=\lim_{\lambda\in\Lambda}f(x_\lambda)=\lim_{\lambda\in\Lambda}g(x_\lambda)=g(x) $$ so that we finally conclude that $f$ and $g$ are equal.

So clearly in the above argumentations I do not use hausdorff-separability but only this relevant result which does not require that $Y$ is hausdorff: however I well know that if $Y$ is not Hausdorff then the limit of $f(x_\lambda)$ and $g(x_\lambda)$ are not generally unique and I realise that this could be a problem but unfortunately I was not able to find a counterexamle showing that the result could not hold when $Y$ is not an hausdoff space and thus I thought to put a specific queston. So could someone help me, please?

Antonio Maria Di Mauro
  • 4,812
  • 2
    If $Y$ has the indiscrete topology, then any function $f: X \rightarrow Y$ is continuous. By this, it is easy to find counter-examples. – Ulli Jun 12 '22 at 08:30
  • @Ulli Sorry, I do not see this. So if $Y$ has the indiscrete topology then $f(x_\lambda)$ and $g(x_\lambda)$ converge to any $y\in Y$, right? – Antonio Maria Di Mauro Jun 12 '22 at 08:36
  • Yes. For instance, let $X = Y$ be a set with more than one point with the indiscrete topology. Consider two functions $f, g: X \rightarrow Y$, which agree on one point, but differ on another point. Or, to have a non-trivial $X$, let $X$ be the reals with the usual topology, $Y$ the reals with the indiscrete topology, and modify the identity at one point. – Ulli Jun 12 '22 at 08:45
  • 1
    @Ulli Sorry, but if $X=Y$ and if $Y$ has the indiscrete topology then the only dense set of $X$ is exactly $X$, right? – Antonio Maria Di Mauro Jun 12 '22 at 08:50
  • If $X$ has the indiscrete topology, any non-empty subset of $X$ is dense in $X$. – Ulli Jun 12 '22 at 09:06
  • @Ulli Sorry, but if $X$ has the indiscrete topology then any subset of $X$ is closed if the indiscrete topology is just the power set of $X$. Perhaps with the indiscrete topology you want refer to the topology whose open sets are $X$ and the empty set? – Antonio Maria Di Mauro Jun 12 '22 at 09:09
  • No, it's not me who want to refer to this definition of indiscrete topology. It's the usual definition. Therefore it's called in... – Ulli Jun 12 '22 at 09:13
  • 2
    The power set topology is the discrete topology! – Ulli Jun 12 '22 at 09:14
  • @Ulli Okay, so in this case effectively any subset of $X$ is dense. – Antonio Maria Di Mauro Jun 12 '22 at 09:15
  • @Ulli So could you explain why if $f$ and $g$ agree in one point then they not necessarily agree to other points? – Antonio Maria Di Mauro Jun 12 '22 at 09:17
  • In my above, first, example you should choose $f, g$ in such a way. – Ulli Jun 12 '22 at 09:20

1 Answers1

3

Consider $X=(\Bbb{R}, \tau_{std}) $ and $ Y=(\Bbb{R},\tau_{ind})$

$\tau_{ind}=\{\emptyset,\Bbb{R}\}$

Consider $f,g: X\to Y$ defined by

$f(x) =1 $ and $g(x) =\begin{cases}1 &x\in \Bbb{Q} \\0 & \text{otherwise}\end{cases}$

Then $f, g$ both are continuous ( as only open sets in $Y$ are $\emptyset$ and $\Bbb{R}$ )and $f=g $ on $\Bbb{Q}$ but $f\neq g$

Sourav Ghosh
  • 12,997