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Let $X$ be a metric space with metric $d$.

  1. Show that $d: X \times X \rightarrow \mathbb R$ is continuous
  2. Let $X'$ denote a space having the same underlying set as $X$. Show that if $d: X' \times X' \rightarrow \mathbb R$ is continuous, then the topology of $X'$ is finer that the topology of $X$.

I know how to approach question 1 (I think I am doing it right):

Let $(x, y) \in X \times X$; We want to prove that given $\epsilon \gt 0$, there exists opensets $x \in U$ and $y \in V$ such that $$d\left(U, V\right) \subset \left(d\left(x, y\right) - \epsilon, d\left(x, y\right) + \epsilon\right)$$ Let $U = B_d(x, \frac \epsilon 2)$ and $V = B_d(y, \frac\epsilon 2)$. From triangle inequality, we have: $\forall (x', y') \in U \times V$: $$d(x,y) - \epsilon\leq d(x, y) - d(x, x') - d(y, y')\leq d(x', y') \leq d(x, x') + d(x, y) + d(y, y') \leq d(x, y) + \epsilon$$ Therefore we have $d\left(U, V\right) \subset \left(d\left(x, y\right) - \epsilon, d\left(x, y\right) + \epsilon\right)$, which means it's continuous.

However, I don't know how to approach the second question. Aren't this trivial given the fact that $X'$ has the same underlying set as $X$? Because then for any opensets in $X$, it is automatically in $X'$. Why $d$ still have to be continuous?

Prosthobotics
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2 Answers2

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Here is another way to approach the first part.

To prove continuity, you can apply the sequential characterization of it.

Precisely, let $(x_{n},y_{n})$ be a sequence in $X\times X$ which converges to $(x,y)\in X\times X$.

Then the following relation holds: \begin{align*} d_{X}(x,y) \leq d_{X}(x,y_{n}) + d_{X}(x_{n},y_{n}) + d_{X}(x_{n},y) \end{align*}

as well as the next relation: \begin{align*} d_{X}(x_{n},y_{n}) \leq d_{X}(x_{n},y) + d_{X}(x,y) + d_{X}(x,y_{n}) \end{align*}

Consequently, due to limit properties and convergence of the sequences $x_{n}$ and $y_{n}$, we conclude that \begin{align*} \begin{cases} \displaystyle d_{X}(x,y) = \lim_{n\to\infty}d_{X}(x,y) \leq \lim_{n\to\infty}d_{X}(x_{n},y_{n})\\\\ \displaystyle \lim_{n\to\infty}d_{X}(x_{n},y_{n})\leq \lim_{n\to\infty}d_{X}(x,y) = d_{X}(x,y) \end{cases} \end{align*}

Finally, due to the squeeze theorem, we conclude that \begin{align*} \lim_{n\to\infty}d_{X}(x_{n},y_{n}) = d_{X}(x,y) \end{align*}

whence the continuity follows.

Átila Correia
  • 15,530
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Answer for 2): Let $U$ be open in $(X,d)$ and $x \in U$. There exists $r>0$ such that $B_d(x,r) \subset U$. Now consider $E=\{(u,v): d(u,v) <r\}$ This is open in $X'\times X'$ with respect to the product topology. This implies that $B_d(x,r)=\{y: (x,y) \in E\}$ is open in $X'$. We have proved that each point $x$ of $U$ belongs to an open set in $X'$ which is contained in $U$. It follows that $U$ is a union of open sets in $X'$. Thus every open set in $(X,d)$ is open in $X'$.

Kavi Rama Murthy
  • 311,013