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The space $(\mathbb{R}^2, d)$ where $d(x,y)=max \{|x_1-x_2|, |y_1, y_2|\}$ for $x=(x_1, y_1)$ and $y=(x_2, y_2)$ $\in X$ is a complete metric space.

Let $X=\{(0,0), (-\frac{1}{8}, 0), (0, -\frac{1}{5})\}$ with the same distance function defined above.

I have shown that $(X,d)$ is a metric space by brute force since I only have 3 elements in the set (listing all the conditions and showing it).

My problem is how will I show that this subset is a complete metric space. How am I going to construct the Cauchy sequences. Or do you have other solution than this?

Thanks a lot.

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Show that any Cauchy sequence is eventually constant. – David Mitra Mar 12 '14 at 10:30
Note that every finite metric space is necessarily complete. – Joshua Pepper Mar 12 '14 at 11:40
@JoshuaPepper Do you have any detailed discussion about this "every finite metric space is necessarily complete"? Thanks a lot. – reymartsalcedo Mar 14 '14 at 8:15
@reymartsalcedo You can see that this is true by proving David Mitra's statement: every Cauchy sequence must eventually be constant. If a sequence is Cauchy, then $\forall \varepsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n,m \geq N$, $|x_n - x_m| < \varepsilon$. However, if you pick $\varepsilon$ smaller than the minimum of the distances between the points of your space (which exists by finiteness, and is non-zero by the axioms of a metric space), your sequence has to be constant after $N$. – Joshua Pepper Mar 14 '14 at 9:47
Thanks @JoshuaPepper .. This will help a lot. – reymartsalcedo Mar 17 '14 at 6:48

The easiest solution is probably to note that $X$ is finite, thus closed in $\mathbb{R}^2$. We then use that a subset of a complete metric space is complete if and only if it is closed.

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Thanks for the answer! :) – reymartsalcedo Mar 14 '14 at 8:17

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