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.

Let $X$ be a metric space with metric $d$. If $\mathcal{T}$ is a topology on $X$ such that the function $d\colon X \times X \to \mathbb{R}$ is continuous, then how to show that $\mathcal{T}$ is finer than the topology induced by the metric $d$?

In other words, how to prove that if $X$ has a metric $d$, then the topology induced by $d$ is the coarsest topology relative to which the function $d$ is continuous?

share|improve this question
How to insert the capital Greek letter tau? –  Saaqib Mahmuud Feb 7 '13 at 10:52
The upper case Greek tau (Τ) is (virtually) indistinguishable from the upper case Latin tee (T), and so $\LaTeX$ has no special code for the character. –  Arthur Fischer Feb 7 '13 at 10:56
@SaaqibMahmuud I guessed you wanted a caligraphic $T$, i.e. $\mathcal{T}$. This is what I normally see used to denote topologies. –  Matt Pressland Feb 7 '13 at 10:57
How to insert this caligraphic T using LATEX? –  Saaqib Mahmuud Feb 7 '13 at 11:28
@SaaqibMahmuud when a page uses MathJax, you can right click on a piece of maths and in there you'll find an option to show the TeX source. –  kahen Feb 7 '13 at 21:05

2 Answers 2

up vote 7 down vote accepted

To show that the $d$-metric topology is coarser than $\mathcal{T}$ we must show that every $d$-open set is $\mathcal{T}$-open. Of course, it really suffices to show that every $d$-ball is $\mathcal{T}$-open (since the $d$-balls form a basis for the $d$-metric topology). To show that the $d$-ball $B ( x , \epsilon )$ is $\mathcal{T}$-open it suffices to find for each $y \in B ( x , \epsilon )$ a $\mathcal{T}$-open neighbourhood $V$ of $y$ such that $V \subseteq B ( x , \epsilon )$.

Hint: By continuity of $d$ for each $\epsilon > 0$ the set $$U_\epsilon = \{ ( u , v ) \in X \times X : d ( u , v ) < \epsilon \}$$ is open in $X \times X$ with respect to the topology $\mathcal{T}$ (or $\mathcal{T} \times \mathcal{T}$, if you will). Recall that the "open rectangles" (i.e., sets of the form $U \times V$ where $U , V \subseteq X$ are $\mathcal{T}$-open) form a basis for the product topology on $X \times X$.

Detail: Given $x \in X$ and $\epsilon > 0$, we want to show for all $y \in B ( x , \epsilon )$ that there is a $\mathcal{T}$-open set $V$ with $y \in V \subseteq B ( x , \epsilon )$. As above the set $U_\epsilon$ is a $( \mathcal{T} \times \mathcal{T} )$-open subset of $X \times X$ and since $y \in B ( x , \epsilon )$ it follows that $( x , y ) \in U_\epsilon$. Then there are $\mathcal{T}$-open $U , V \subseteq X$ such that $$ ( x , y ) \in U \times V \subseteq U_\epsilon.$$ In particular $V$ is a $\mathcal{T}$-open neighbourhood of $y$. Note that given any $z \in V$ we have that $( x , z ) \in U \times V \subseteq U_\epsilon$, and so by definition of $U_\epsilon$ it follows that $d ( x , z ) < \epsilon$, meaning that $z \in B ( x , \epsilon )$. We may then conclude that $V \subseteq B ( x , \epsilon )$.

share|improve this answer
So far so good! But what next? We know that there are open sets $U$ and $V$ containing $x$ and $y$, respectively. Now we need to show that $V$ is contained inthe ball. But how to? –  Saaqib Mahmuud Feb 7 '13 at 11:35
@SaaqibMahmuud: By our choice of $U$ and $V$ we know something about the distance between any point in $U$ and any point in $V$. $U$ was also chosen to contain a particular (and important) point. –  Arthur Fischer Feb 7 '13 at 11:38
So if $x^\prime \in U$ and $y^\prime \in V$, then we must have $$ d(x^\prime, y^\prime) \leq d(x^\prime,x) + d(x,y) + d(y,y^\prime) < d(x^\prime,x) + \epsilon + d(y,y^\prime). $$ What next? –  Saaqib Mahmuud Feb 7 '13 at 15:52
@SaaqibMahmuud: What is $U \times V$ a subset of? –  Arthur Fischer Feb 7 '13 at 15:56
It is of course a subset of $X \times X$. Can we be any more precise? –  Saaqib Mahmuud Feb 7 '13 at 20:51

The topology induced a collection of functions $\mathcal{F}$ is defined as the coarsest topology in which all of the functions $f\in\mathcal{F}$ are continuous.

In other words the topology induced by $d$ is the same thing as the coarsest topology relative to which the function $d$ is continuous.

$\tau$... \tau... is going to be finer than this topology by definition.

share|improve this answer
Thanks, but I'd like a proof without employing this "definition." –  Saaqib Mahmuud Feb 7 '13 at 11:37
Obviously the OP is not employing this definition of the induced topology, so this is not at all a proof! Maybe you can rewrite it? Thanks. –  awllower Nov 15 '13 at 5:49
@Jp McCarthy you are misunderstanding concepts. The topology induced by a collection of functions $\mathcal{F}$ is a topology defined on the domain of the functions. In this context the top. induced by the metric $d:X\times X\to \mathbb{R}$ (or the family $\mathcal{F}=\lbrace d\rbrace$) is the coarsets top. on $X\times X$ s.t. $d$ is continuous. But that topology need not be related with the product topology of the topology induced by the metric $d$ on $X$. Consider the metric $g$ on $\mathbb{R}$, $g(x,y)=|x-y|$. The coarsest top. on $\mathbb{R}^{2}$ s.t. g is continuous is not the Euclidean. –  Chilote Sep 7 at 1:15

Your Answer


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.