**Question:**

\(X\) is a metric space. \(Y\) is a closed subset of \(X\) such that the distance between any two points in \(Y\) is at most 1. Then

A. \(Y\) is compact.

B. any continuous function from \(Y\to \mathbb{R}\) is bounded.

C. \(Y\) is not an open subset of \(X\)

D. none of the above.

**Discussion:**

Let \(X=\) an infinite set for example \(=\mathbb{R}\) with the metric as discrete metric.

That is \(d(x,y)=1\) if \(x\neq y\) and \(d(x,y)=0\) if \(x=y\).

Then every set in \(X\) is open and every set is closed.

Now take \(Y=X\). Then \(Y\) can be covered by singleton sets. \(Y=\cup \{\{a\}|a\in Y\}\). Now each of the singleton sets is open in discrete metric space. Therefore, this is an open cover for \(Y\). Since \(Y\) is infinite, this cover has no finite subcover. So \(Y\) is **not compact.**

Given any \(f:Y\to \mathbb{R}\), for open set \(U\in \mathbb{R}\), \(f^{-1}(U)\subset Y\). Since Y is discrete, \(f^{-1}(U)\)is open in \(Y\). So every function \(f:Y\to \mathbb{R}\) is a continuous function. In particular if we define \(f(x)=x\) then \(f\) is a continuous function. And \(f\) is **not bounded.**

Also, every subset of \(X\) is open. So \(Y\) **is open**.

Therefore, we are left with **none of the above.**

## No comments, be the first one to comment !