Question

Which one of the following options is necessarily true?

• A. Every contraction map on a metric space is uniformly continuous.
• B. The image of an open set under a continuous function is an open set.
• C. Every complete metric space is compact.
• D. Every metric space is second countable.

Hint:

A contraction map is Lipschitz continuous, and every Lipschitz map is uniformly continuous.

Answer 1

The Correct Option is A

Step 1: A contraction map \(f:X\to X\) on a metric space \((X,d)\) satisfies \(d(f(x),f(y))\le k\,d(x,y)\) for all \(x,y\in X\), for some constant \(0\le k<1\). This is exactly the statement that \(f\) is Lipschitz continuous with Lipschitz constant \(k\).

Step 2: Every Lipschitz map is uniformly continuous: given \(\varepsilon>0\), if \(k>0\) choose \(\delta=\varepsilon/k\) (if \(k=0\), \(f\) is constant and trivially uniformly continuous); then \(d(x,y)<\delta\) gives \(d(f(x),f(y))\le k\,d(x,y)<k\delta=\varepsilon\). This single \(\delta\) works for every pair \(x,y\) in \(X\), regardless of location, so \(f\) is uniformly continuous. Option (A) is necessarily TRUE for every contraction map on every metric space.

Step 3: Option (B) is false in general: let \(f:\mathbb{R}\to\mathbb{R}\) be the constant map \(f(x)=c\). This is continuous, but the image of the open set \(\mathbb{R}\) is the single point \(\{c\}\), which is not open in \(\mathbb{R}\). Continuous maps need not be open maps.

Step 4: Option (C) is false: \(\mathbb{R}\) with the usual metric is complete (every Cauchy sequence converges), but it is not compact, since it is unbounded.

Step 5: Option (D) is false: let \(X\) be an uncountable set with the discrete metric \(d(x,y)=1\) for \(x\ne y\) and \(d(x,x)=0\). Every singleton \(\{x\}\) is open (it is the ball of radius \(\tfrac12\) around \(x\)), and any basis for the topology must include a set contained in \(\{x\}\) for every \(x\), forcing the basis to contain uncountably many singletons. So \(X\) has no countable base, and is not second countable.

\[\boxed{\text{Option (A): Every contraction map on a metric space is uniformly continuous}}\]