# Topological space

• May 1st 2010, 01:30 AM
johndon
null
removed.
• May 1st 2010, 11:13 AM
nimon
I'm not sure what you mean by a 'net', but I can show you $\displaystyle A \Leftrightarrow C$ to get you started.

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

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

Hope this helps.
• May 2nd 2010, 08:23 PM
Drexel28
Quote:

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.

Quote:

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

Quote:

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