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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. – arjafi 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. – Matthew 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
up vote 9 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 $V \subseteq B ( x , \epsilon )$.

First, the continuity of $d$ implies that for each $\epsilon > 0$ the set $$U_\epsilon := \{ ( u , v ) \in X \times X : d ( u , v ) < \epsilon \} = d^{-1} [ ( -\infty , \epsilon ) ]$$ is open in $X \times X$ with respect to the topology $\mathcal{T}$ (or $\mathcal{T} \times \mathcal{T}$, if you will).

Now, 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 $y \in B ( x , \epsilon )$, it follows that $( x , y ) \in U_\epsilon$, and since $U_\epsilon$ is a $( \mathcal{T} \times \mathcal{T} )$-open subset of $X \times X$, 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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I think a consequence of the reasoning behind part of this proof is that each "slice" of an open set in a product space $X\times Y$ is open. To wit, if $W$ is open in $X\times Y$, then for $x\in X$, the slice defined by $S_x(W)=\{y\in Y:(x,y)\in W\}$ is open in $Y$. Of course if this result was known beforehand then part of this proof could be sped up as well. – Fang Jing Apr 4 '14 at 21:38

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