1. ## null

removed.

2. I'm not sure what you mean by a 'net', but I can show you $A \Leftrightarrow C$ to get you started.

First, if $X$ is Hausdorff then, given any $(x,y)\in (X\times X) \backslash \Delta$ we can find disjoint, open neighbourhoods $U_{x},U_{y}\subseteq X$ of $x$ and $y$ respectively. Then $(x,y)\in U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta$ (the last inclusion is due to the disjointness of the neighbourhoods: if $(z,z) \in U_{x}\times U_{y}$ then $z\in U_{x} \cap U_{y}$ which is wrong by assumption). It follows that $(X\times X)\backslash \Delta$ is a union of open sets in $X \times X$, so must be open, and therefore $\Delta$ is closed.

The converse is similar. If $\Delta$ is closed then $(X\times X)\backslash \Delta$ is open, and so for any $(x,y)\in X\times X$ we can find an open set $V\subseteq (X\times X) \backslash \Delta$ such that $(x,y)\in V$, and it follows that $(x,y)\in U_{x}\times U_{y} \subseteq V$ for some $U_{x},U_{y}$ open in $X$. Since $U_{x}\times U_{y} \subseteq (X\times X) \backslash \Delta$, they must be disjoint, and so $X$ is Hausdorff.

Hope this helps.

3. Originally Posted by johndon
Q(i) Let X be a topological space. Show that the following
statements are equivalent.
(A) X is a Hausdorff space.
(B) Every net in X has at most one limit.

(C) The diagonal delta= {(x,x):x∈X }⊂ X x X is closed in X × X

(ii) Let X,Y be topological spaces, A ⊂ X, and Ᾱ = X. Let f: X →Y ,
g: X →Y be continuous functions, and let f(a)= g(a) ∀ a∈A . Show that if
Y is a Hausdorff space, then f(x)= g(x) ∀ x∈X . [HINT: Use (B) or (C).]
Hint:

Spoiler:

The set of those points is closed.

Partial solution (only look at this after much, much thought. Even then, feel bad):
Spoiler:

Let $\varphi,\psi:X\to Y$ be continuous and let $A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}$. Now, define another map $\varphi\oplus\psi:X\to Y\times Y:x\mapsto(\varphi(x),\psi(x))$, this is easily verifed to be continuous and $A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}\left(\Delta_Y\right)$ from where the conclusion follows