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Question: show that the metric space $(\mathbb{R^n}, d)$ is connected. Where $d$ is maximum metric (chebyshev distance)

Given Hint: every non-empty proper subset of $\mathbb{R^n}$ has a non-empty boundary.

My attempt: I know $\mathbb{R^n}$ with a usual distance metric is connected. Further I know that, these two metrics (usual metric and maximum metric) on $\mathbb{R^n}$ are equivalent and hence $\mathbb{R^n}$ with respect to maximum metric is connected.

I don't know my attempt is correct or not. Further I like two know if the two metrics are equivalent then what can we say? I didn't used the hint! How can i use it?

Please help.

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Indeed you have the right idea.

$(\Bbb{R}^k ,d_2)$ is homeomorphic to $(\Bbb{R}^k ,d_{\infty})$ since the two metrics are equivalent.

You do not need to use the hint at all.

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The hint doesn't make much sense: for a topological space, being connected is equivalent to the property that every non-empty proper subset has a non-empty boundary. (A clopen set has en empty boundary. For the converse, if $A$ is a proper nonempty subset with empty boundary, then it is not dense, and its closure also has empty boundary. For a closed set, having an empty boundary is equivalent to being clopen.)

You argument is sound (once you know that equivalence of metrics implies homeomorphism, and connectedness is preserved by homeomorphisms).

For a more general argument, note that a that for a real vector space $V$ to be (path!) connected, it is sufficient for the multiplication to be left-continuous, i.e. that for every $v$, the map $\mathbf R\to V$ given by $t\mapsto tv$ is continuous. This is true for pretty much any reasonable topology on a real vector space.

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Connectedness only depends on the topology and both metrics induce the same topology on $\Bbb R^k$, namely the product topology. Products of connected spaces are connected.

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