# Some questions concerning continuity and relations

A lot of equivalent conditions for functions between topological spaces $$X\overset f\longrightarrow Y$$ are proved on this site. Here some of them formulated from the perspective of 'relations':

$(1)\quad$For all sets $M\subset Y$, $(x,y)\in f \wedge x\in \overline{f^{-1}(M)}\Rightarrow y\in \overline{M}$

$(2)\quad$For all sets $M\subset Y$, $\overline{f^{-1}(M)}\subset f^{-1}(\overline{M})$

$(3)\quad$$V\subset Y is open \Rightarrow f^{-1}(V)\subset X is open (4)\quad$$F\subset Y$ is closed $\Rightarrow f^{-1}(F)\subset X$ is closed

$(5)\quad$For all sets $L\subset X$, $f(\overline{L})\subset\overline{f(L)}$

Consider the formulas as strings of symbols.

Which pairs of the strings corresponds to equivalent conditions, if $f\subset X\times Y$ is not a function, but a:

$\;$A.$\,$ partial function (a function only defined on a part of $X$)?

$\;$B.$\,$ multivalued function?

$\,$C.$\,$ relation in general?

$\,$D.$\,$ special sort of relations (a sort of your choice)?

Example: Define the relation $f\subset\mathbb{R}\times\mathbb{R}$ by $(x,y)\in f\Leftrightarrow x^2+y^2=1$. For the relation $f$ the conditions $(1)$, $(2)$, $(4)$ and $(5)$ still holds, but $(3)$ do not, so the condition $(3)$ is not equivalent with any of the others, in the extended sense.

• – Travis Willse Aug 27 '14 at 8:59
• The question might be of interest for 'continuous relations' but is interesting in itself. – Lehs Aug 27 '14 at 9:19

None of $(1)$-$(5)$ seems to be equivalent, except $(2)$ and $(4)$.
First, $(x,y)\in f\Leftrightarrow x^2+y^2=1$ satisfies all conditions but $(3)$, why $(3)$ isn't equivalent with any of the others.
Second, $(1)$ differs from the others for $f=X\times Y$. The conditions $(2)$-$(5)$ are satisfied by $X\times Y$, while that relation generally not satisfy $(1)$.
Third, $(x,y)\in f\Leftrightarrow x^2+y^2=1\wedge x<1$ (a dotted circle) satisfy $(5)$ but not $(2)$ or $(4)$.
Last, $(2) \Rightarrow (4)$ (select M closed). Reverse: suppose $(4)$ and $M\subset Y$. Then
$\overline{f^{-1}(M)}\subset \overline{f^{-1}(\overline M)}= f^{-1}(\overline M)$.