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I have recently started reading Kosniowski's A First Course in Algebraic Topology; my background on Calculus, Algebra and Set Theory is vast enough to understand many of the concepts shown in the first few pages. This is my first time exploring an entire subject on my own, and the lack of a solution manual is overwhelmingly inconvenient.

Let $(A, d)$ be a metric space with metric $d$. Let $y \in A$. Show that the function $f: A\rightarrow \mathbb{R}$ where $f(x) = d(x, y)$ is continuous, where $\mathbb{R}$ has the usual metric (which we`ll call $d_\mathbb{R}(x_0, y_0)$).

My geometrical insight:

The function $f(x)$ is the distance between the point $x$ we want to show continuity on and some other fixed point $y$. Then, let $a, b \in A$; draw two circumferences (or higher-dimensional equivalents) centered in $y$, respectively passing through $a$ and $b$. The expression: $$d_\mathbb{R}(f(a), f(b))$$ Is the difference of the two radii.

The definition of continuous function at $x$ is given as: $$\forall \epsilon \in \mathbb{R}_{>0}: \exists \delta \in \mathbb{R}_{>0}: \forall x \in A_1: d_1 \left({x, a}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({a}\right)}\right) < \epsilon $$ My attempt:

  • Let $a, b \in A$. Then: $$d_\mathbb{R}(f(a), f(b)) < \epsilon$$ $$-\epsilon < f(a) - f(b) < \epsilon$$ $$-\epsilon + 2f(b) < f(a) + f(b) < \epsilon + 2f(b)$$ $$-\epsilon + 2d(b, y) < d(a, y) + d(b, y) < \epsilon + 2d(b, y)$$ By the triangular inequality: $$d(a, b) \le d(a, y) + d(b, y) < \epsilon + 2d(b, y)$$ $$d(a, b) < \epsilon + 2d(b, y)$$ And given the result above, a good candidate for $\delta$ might be: $$\delta = \epsilon + 2d(b, y)$$ But I feel I should be thinking the other way round, and this approach could be completely wrong. I don't even know an effective method of disproving myself, or checking if what I've found actually works...
  • An idea stemming from the first comment and the first answer led me to this reasoning: $$ - \epsilon < d(a,y) - d(b,y) < \epsilon$$ $$ d(a,y) - d(b,y) \le d(b,a) < \delta $$ And it works symmetrically to get: $$ -\delta < - d(b,a) \le d(a,y) - d(b,y) $$ Then, a good candidate for $\delta$ is: $$ \delta \le \epsilon $$ Applying it to a concrete situation (I tried with both metric spaces being the real number line with the usual metric) seemed to work.
Niki Di Giano
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  • I think you've got $a$ and $b$ the wrong way round in defining your candidate - you want a $\delta$ depending on $a$ and $\epsilon$ such that for any $b$, if $d(a, b) < \delta$ then ... One way of testing your ideas is to apply them to a concrete example. Try $(A, d) = (\Bbb{R}, d_{\Bbb{R}})$, $y = 0$, $a = 10$ and $\epsilon = 1/4$. I think you'll find your candidate is much too big. – Rob Arthan Sep 14 '17 at 19:59
  • I have uploaded a new attempt based on yours and the first answer's advice. – Niki Di Giano Sep 14 '17 at 21:14

1 Answers1

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For metric spaces $(A,d)$ and $(B,e),$ a necessary and sufficient condition that $f:A\to B$ is continuous is that whenever $x\in A$ and $(x_n)_n$ is a sequence in $A$ with $\lim_{n\to \infty}d(x_n,x)=0$ then $\lim_{n\to \infty}e(f(x_n),f(x))=0.$

In your Q let $x\in A$ and let $(x_n)_n$ be a sequence in $A$ with $\lim_{n\to \infty}d(x_n,x)=0 .$ We have $|f(x)-f(x_n)|\leq d(x,x_n)$ so $\lim_{n\to \infty}d_{\mathbb R}(f(x_n),f(x))=0.$

The inequality $|f(x)-f(x_n)|\leq d(x,x_n)$ is obtained as follows:

(1). $d(x,y)\leq d(x,x_n)+d(x_n,y)$ so we have $$(*)\quad f(x)-f(x_n)=d(x,y)-d(x_n,y)\leq d(x,x_n).$$ (2). $d(x_n,y)\leq d(x_n,x)+d(x,y)$ so we have$$(**)\quad f(x_n)-f(x)=d(x_n,y)-d(x,y)\leq d(x_n,x)=d(x,x_n).$$ From $(*)$ and $(**)$ we have $|f(x)-f(x_n)|\leq d(x,x_n).$

Remark: In a metric space we have $|d(a,c)-d(b,c)|\leq d(a,b)$ so we can obtain $|f(x)-f(x_n)|\leq d(x,x_n)$ in one step rather than by combining the steps $(*)$ and $(**)$.

DanielWainfleet
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  • Thanks for your answer, which is really educative and precise. The statement given in the beginning was really helpful, although it has not been mentioned in the book so far. Is it a generalization of the convergent sequence definition for limits in $\mathbb{R}^n$, applied to the definition of continuity? – Niki Di Giano Sep 14 '17 at 20:54
  • Yes it is that generalization. I am not familiar with that book. I think it is best to start with topological spaces in general and (general) continuity and some other basic topics concerning all spaces, and only then move to the topic of metric spaces. – DanielWainfleet Sep 14 '17 at 21:15