Could someone please check my proof that $(x_n) \text{ is Cauchy in } X\implies (f(x_n)) \text{ is Cauchy in } Y$

Question

Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $f:X\to Y$.

Show that if $f:X\to Y$ is uniformly continuous, then

$$(x_n) \text{ is Cauchy in } X\implies (f(x_n)) \text{ is Cauchy in } Y$$

My working:

Since $f$ is uniformly continuous then $\forall\epsilon>0, \exists\delta>0$ s.t. $\forall x,x'\in X, d(x,x')<\delta\implies\rho(f(x),f(x'))<\epsilon$

But since $(x_n)$ is Cauchy in $X$, then $\forall\epsilon>0, \exists N$ s.t. $\forall m,n\geq N\implies d(x_m,x_n)<\epsilon$.

Then $\forall\epsilon>0,\exists\delta>0$ s.t. $\exists N$ s.t. $\forall m,n\geq N\implies d(x_m,x_n)<\delta\implies\rho(f(x_m),f(x_n))<\epsilon$.

So we have shown that $\forall\epsilon>0, \exists N$ s.t. $\forall m,n\geq N\implies \rho(f(x_m),f(x_n))<\epsilon$.

Hence $(f(x_n)) \text{ is Cauchy in } Y$.

Could anybody please check my working whether it is correct? If so, how can I write in a better way? If it is wrong, where did I make the mistake?

Thanks for the help!

Answer 1

I don't intend to write a completely new proof because your argument looks fine to me but I think that you can write the proof in a more systematic way. Although I think that your essential idea is correct, it is better to emphasize a few subtleties in the writing of the proof.

Since $f$ is uniformly continuous then $$∀ϵ>0,∃δ>0\mid ∀x,x′∈X,d(x,x′)<δ⟹ρ(f(x),f(x′))<ϵ\tag{1}$$ But since $(x_n)$ is Cauchy in $X$, then $$∀ϵ>0,∃N\mid ∀m,n≥N⟹d(x_m,x_n)<\epsilon\tag{2}$$

Now let us choose $\epsilon>0$. Then by $(1)$ $\color{red}{\text{for our particular choice of}}$ $\epsilon>0$ we can write, $$∃δ>0\mid ∀x,x′∈X,d(x,x′)<δ⟹ρ(f(x),f(x′))<ϵ\tag{3}$$ Now let us choose any $\delta>0$ which satisfies $(3)$. Then for $\color{red}{\text{that}}$ $\delta>0$ we will have by $(2)$, $$∃N\mid ∀m,n≥N⟹d(x_m,x_n)<\delta\tag{4}$$Now by $(1)$ $\color{red}{\text{for our particular choice(s) of}}$ $\epsilon$ and $\delta$ we have, $$\exists N\mid (\forall m,n\ge N)\land(\forall x_m,x_n\in X), d(x_m,x_n)<\delta\implies \rho(f(x_m),f(x_n))<\varepsilon$$ Since $\epsilon$ was arbitrary so we can write, $$∀ϵ>0,(∃δ>0)\land(∃N) \mid ∀m,n≥N⟹d(x_m,x_n)<δ⟹ρ(f(x_m),f(x_n))<ϵ$$ So we have shown that $$∀ϵ>0,∃N\mid ∀m,n≥N⟹ρ(f(x_m),f(x_n))<ϵ$$ Hence $(f(x_n))$ is Cauchy in $Y$.

Answer 2

the result is correct but in the demonstration it is best not to confuse the symbol $\varepsilon$ and $\delta$

Since $f$ is uniformly continuous then $\forall\epsilon>0, \exists\delta>0$ s.t. $\forall x,x'\in X, d(x,x')<\delta\Rightarrow\rho(f(x),f(x'))<\epsilon$

But since $(x_n)$ is Cauchy in $X$, then $\forall\delta>0, \exists N$ s.t. $\forall m,n\geq N\Rightarrow d(x_m,x_n)<\delta$.

Then $\forall\epsilon>0,\exists\delta>0$ s.t. $\exists N$ s.t. $\forall m,n\geq N\Rightarrow d(x_m,x_n)<\delta\Rightarrow\rho(f(x_m),f(x_n))<\epsilon$.

So we have shown that $\forall\epsilon>0, \exists N$ s.t. $\forall m,n\geq N\Rightarrow \rho(f(x_m),f(x_n))<\epsilon$.