Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On page 85 of the book it reads the following definition:

Definition 4. Suppose that for each pair of distinct points $x$ and $y$ in a topological space $T$, there is a neighborhood $O_x$ of $x$ and a neighborhood $O_y$ of $y$ such that $x \in O_y, y \in O_ x$. Then $T$ is said to satisfy the first axiom of separation, and is called a $T_1$- space.

But in other sources, such as Mathworld, they define

$T_1$-separation axiom : For any two points $x,y \in X$ there exists two open sets $U,V$ such that $x\in U$ and $y \notin U$ , and $y\in V$ and $x \notin V$.

Are both definitions equivalent?

share|improve this question
4  
Must be a typo. Certainly $x\notin O_y$ and $y\notin O_x$ was meant. (Note as written, every space satisfies the condition. Take $O_x=O_y=T$.) –  David Mitra Jan 21 '13 at 22:33

2 Answers 2

up vote 4 down vote accepted

It's certainly a typographical error. "$x\notin O_y$" and "$y\notin O_x$" was meant instead of "$x\in O_y$" and $y\in O_x$". (This is used in the Theorem immediately following the definition).

With this correction, the two notions coincide (trivially).

But to answer you directly, note that every topological space $T$ satisfies the criteria as written: just take $O_y=O_x=T$.

share|improve this answer

Google

Kolmogorov Fomin errata

and find my errata sheets still out there on the internet. (Yes, this one is listed.) K & F is a good book for an introduction to the topics, plus it has a great price.

share|improve this answer
    
Thank you for creating this errata! You should link it on your homepage so google search can find it. I had this exact problem reading this book today. –  integrator 2 days ago

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.