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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?

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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
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2 Answers

up vote 5 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 $V \subseteq B ( x , \epsilon )$.

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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
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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.

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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
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