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Heads up: I am very new to abstract algebra and proofs.

Take the non-negative real line $X = 0 \cup \mathbb{R}^+=[0, +\infty)$. We know that $X\subset \mathbb{R}$ and that $\mathbb{R}$ is complete. Define now the Euclidean metric $d(x,y) = \lvert x - y \rvert$. Then $(X,d)$ spans a metric space. I wish to show that this metric space is complete.

So for an arbitrary sequence $(x_n)$ we need $d(x_n, x) < \varepsilon$, where $x\in X$, that is the sequence has to converge to $x$ which is in the set we are working with. Then $(x_n)$ is Cauchy, and since the sequence was arbitrary $X$ is complete.

This is what I know we require, however, I am unsure if I am on the right track.

My attempt:

Consider a sequence that is Cauchy, $(x_m)$. Then for every $\varepsilon$, there is an $N=N(\varepsilon)$ such that $$d(x_m, x_r) = \lvert x_m - x_r\rvert < \varepsilon .$$ As $m\to \infty$ $x_m\to x$. How do I prove that $x\in X$ so that the arbitrary Cauchy sequence converges to a point in $X$ (thereby completing the proof)?

Any help would be appreciated.

José Carlos Santos
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Sindre Bakke Øyen
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1 Answers1

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Since $[0,+\infty)$ is a closed subset of $\mathbb R$, if $(x_n)_{n\in\mathbb N}$ is a sequence of elements of $[0,+\infty)$ which converges to a real number $x$, then $x\in[0,+\infty)$.

José Carlos Santos
  • 427,504