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.

From Mendelson's Introduction to Topology:

Let $(X,d)$ be a metric space. ... [Then] $X$ is open.

But consider the closed interval $[a,b]$ over the real numbers. Presumably there is a distance function $d$ such that $([a,b],d)$ is a metric space. But $[a,b]$ is not an open set. What, then, have I missed?


To expand upon my question, here is my rough counterproof.

An open set is a neighborhood of each of its points. Consider the metric space $(X,d)$, with $X = [a,b]$ and $d(x,y) = \sqrt{x^2+y^2}$. If $X$ is open, then $X$ must be a neighborhood of $a$, which means there must be an open ball $B(a;\delta)\subset X$ for some $\delta > 0$. Yet if $\delta > 0$, then there is some $c \in B(a;\delta)$ such that $c < a$. Therefore $c \notin X$; a contradiction.

What is the flaw in my logic here?

share|improve this question
    
$[a,b]$ is open in itself. Any metric space is its own universe. By definition, any open ball is contained in the universe. –  Prahlad Vaidyanathan Sep 22 '13 at 14:55
    
$[a,b]$ is not open in $\mathbb R$, but it is open in $[a,b]$ itself. –  Santiago Canez Sep 22 '13 at 14:57
    
Think about this; $R$ is not open in $R^2$, but it is open in $R$. open/closed property is very dependent on the underlying space. (anyway how can I write the real number set R? mathds doesnt work... $\mathds{R}$) –  dust05 Sep 22 '13 at 15:01
    
Thanks all for your comments - I have expanded my question to clarify this. @dust05: \mathbb{R} –  Doubt Sep 22 '13 at 15:17
    
I think the natural definition of the open ball in $X$ is defined as $B(x;\delta)\cap X$ in this case. –  dust05 Sep 22 '13 at 15:22
show 1 more comment

1 Answer

up vote 2 down vote accepted

Any metric space is a topological space and the space itself is always open with respect to its topology. It may fail to be open with respect to the topology of some other topological space it is a subspace of. Here, no such super-space is mentioned.

Or reread the definiion of open set exactly: If $(X,d)$ is a metric space, then for $x\in X$ and $r\in \mathbb R_{>0}$ the set $B_r(x):=\{\,y\color{red}{\in X}\mid d(x,y)<r\,\}$ is called open $r$-ball around $x$ and $U\subseteq X$ is called open iff for each $x\in U$, there exists some $r>0$ with $B_r(x)\subseteq U$. For $X=[0,1]$ and the usual metric note that these definition make e.g. $B_{\frac12}(\frac13)=[0,\frac56)$ an open set.

share|improve this answer
add comment

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.