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4.6 Continuity and closedness

We have already studied the relation between convergence and closedness, also, we studied the relation between continuity and convergence. Now, we will see how the notions of continuity and closedness relate to each other.

Proposition 4.6.1

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function and $C\subset\mathbb{R}$ be a closed set. Let $f^{-1}(C)=\{x\in\mathbb{R}\,\mid\,f(x)\in C\}\,.$ Then the set $f^{-1}(C)$ is closed.

Proof: We proceed by contradiction. Suppose that the set $f^{-1}(C)$ is not closed. Then there exists a sequence $\{x_{n}\}^{\infty}_{n=1}\subset f^{-1}(C)$ converging to $x$ , such that $x\notin f^{-1}(C)$ . That is, $f(x)\notin C$ . By our assumption, for any $n\geq 1$ , $f(x_{n})\in C$ . So, by continuity, $\lim_{n\to\infty}f(x_{n})=f(x)\in C$ , leading to a contradiction. $\square$

Remark 4.6.1

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for any non-zero integer $n$ , $f(n)=\frac{1}{n}$ and $f(0)=1$ . Then $\mathbb{Z}$ is a closed subset of $\mathbb{R}$ , and $f(\mathbb{Z})$ is not a closed set.

On the other hand, we have the following proposition on the continuous images of closed sets.

Proposition 4.6.2

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and $C\subset[0,1]$ be a closed subset. Then $f(C)$ is closed as well.

Proof: Suppose that $f(C)$ is not closed. That means that there exists a sequence $\{x_{n}\}^{\infty}_{n=1}\subset C$ , so that $f(x_{n})\to y$ and $y\notin f(C)$ . Our problem is that the sequence $\{x_{n}\}^{\infty}_{n=1}$ is not necessarily convergent. As they did many times, Mr. Bolzano and Mr. Weierstrass help us out! There exists a subsequence (now we use the fact that we defined the function $f$ on a bounded interval, not on the whole real line) $\{x_{n_{k}}\}^{\infty}_{k=1}$ converging to $x$ . Since $C$ is closed, $x\in C$ . By continuity, $f(x)=y$ , hence $y\in f(C)$ , leading to a contradiction. $\square$