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let the standard metric $d(x,y)=|y-x|$ and $\rho_{disc}$ be the discrete metric defined by

$${\displaystyle \rho (x,y)={\begin{cases}1&{\mbox{if}}\ x\neq y,\\0&{\mbox{if}}\ x=y\end{cases}}}$$ Let $V=(\mathbb{R},\rho_{disc}$) and $W=(\mathbb{R},d)$ be two metric spaces (that has $\mathbb{R}$ as underlying sets, but given by two different metrices as written).

  1. Show that every function $h:V\rightarrow W$ is continuous
  2. Show that a function $f:W\rightarrow V$ is continuous if and only if it is constant.

So far

1)

I have a result (though I have to explain it) that states in a discrete metric space that all subsets are open. This implies that if a set $X$ has a discrete metric then all mappings $f:X\rightarrow Y$ to some set $Y$ are continuous. In this case $V$ has a discrete metric and thus it is sufficient to argue that in a discrete metric space that all subsets are open.

We can translate the issue into a ball has radius $r$. If $r<1$ then the only point it contains is the point it's centred on. So any single point has a ball of some radius around it containing only that point. This is the same thing as $B_{0<r<1}(x)=\{x\}$, so we know that every singleton is open.

2)

This one I do not really know how to approach and the fact that it is a bi implication does not help me.

Asaf Karagila
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  • (2) Take a point $y$ in the range of $f$. This is $f^{-1})({y})$ is non-empty. If $f$ is continuous, and since ${y}$ is open, we must have that $f^{-1}({y})$ is open. Since ${y}^{c}$ is also open, then $f^{-1}({y}^c)=(f^{-1}({y}))^c$ is also open. Since $(\mathbb{R},d)$ is connected and since $f^{-1}({y})$ is non-empty, we must have $f^{-1}({y}^c)$ empty. Therefore $f\equiv y$. – plop Jun 14 '21 at 14:38
  • @plop I was told that I could use the supremum property on R directly or indirectly. –  Jun 14 '21 at 15:28
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    Yes, you can use it to replace the knowledge that $(\mathbb{R},d)$ is connected. You could argue like this: We already have that $A=f^{-1}({y})$ is open and non-empty. Pick $h\in A$. If the set of $L= {x\in f^{-1}({y}^c\mid\ x <h)$ is non-empty, then it would have a supremum. This supremum cannot be in $L$, since $L$ is open, but it cannot be in $A$ either, since $A$ is open. Therefore $L$ is empty. Next consider the set $U={x\mid -x\in f^{-1}({y}^c),\ -x>h}$. If $U$ is non-empty, then it has a supremum. – plop Jun 14 '21 at 15:42
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    Again, since $U$ is open and since $A$ is open, then that supremum cannot be in $U$ nor in $A$. Therefore, $U$ is empty. It follows that $f^{-1}({y}^c)$ is empty and therefore, $f\equiv y$. – plop Jun 14 '21 at 15:43
  • @plop Is $A$ the one with the discrete metric? The notation is not that obvious when we define the standard metric as $d(x,y)=|y-x|$ –  Jun 17 '21 at 15:33
  • $A$ is a subset of your $W$. – plop Jun 17 '21 at 16:30

2 Answers2

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Hint: Let $f: W \to V$ be continuous. Let $\varepsilon = 1/2$. There exists some $\delta > 0$ such that if $|x-y| < \delta$ then $\rho(f(x),f(y)) < 1/2$ which implies that $f(x) = f(y)$. This shows that $f$ is piecewise contstant. Can you finish it from here?

George Giapitzakis
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  • I am the absolut worst at $\epsilon$-style proofs. –  Jun 14 '21 at 14:46
  • I was told that I could use the supremum property on $\mathbb{R}$ directly or indirectly. –  Jun 14 '21 at 15:12
  • @user879295 What I would do is take some $x,y \in \mathbb{R}$ and try to show that $f(x)=f(y)$ by partitioning their euclidean distance appropriately. – George Giapitzakis Jun 14 '21 at 15:32
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Hint: In general, a continuous non-constant function from a space $X$ to $V$ exists if and only if $X$ is a disconnected space.

Ben Grossmann
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