Continuous Function Iff Set is Open
Prove $f$ is continuous iff for every closed set $K$ in $\mathbb R$, the set $f^{-1}(K)$ is closed in $\mathbb R$
Solution 1
By definition, $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous if for every open set $O \subseteq \mathbb{R}$, $f^{-1}(O)$ is open.
Let $K \subseteq \mathbb{R}$ be a closed set. Then $K^{c} = \mathbb{R} \setminus K$ is open.
$f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous if and only if $f^{-1}(K^{c})=\mathbb{R} \setminus f^{-1}(K)$ if and only if $f^{-1}(K)$ is closed.
Solution 2
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Suppose that $f$ is continuous. Let $\{x_n\} \subset f^{-1}(K)$ with $x_n \to x$ as $n \to \infty$. Our goal is to prove that $x\in f^{-1}(K)$. It is easy to see that $f(x_n) \subset K$ and $f(x_n) \to f(x)$ as $n\to \infty$. Since $K$ is closed, then $f(x)\in K$, and hence $x\in f^{-1}(K)$.
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Now suppose that for every closed $K$, $f^{-1}(K)$ is also closed. Our goal is to prove that $f$ is continuous, i.e., for any $\{x_n\}$ with $x_n \to x$ as $n \to \infty$, $f(x_n)\to f(x)$ as $n\to \infty$. Let $K= \overline{\{f(x_n)\}}$. Clearly, $K$ is closed. Then $ f^{-1}(K)$ is closed. If $f(x) \not=\lim_{n\to \infty}f(x_n)$, then $f(x) \notin K=\overline{\{f(x_n)\}}$, then $x \notin f^{-1}(K)$. However $\{x_n\}\subset f^{-1}(K)$, which shows that $x \in f^{-1}(K)$. This is a contradiction!
Solution 3
Yet another solution using $\epsilon$-$\delta$.
Suppose $f$ is continuous, $U\subset \mathbb{R}$ is open and $a\in f^{-1}(U)$. Since $f(a)\in U$ and $U$ is open, we can find $\epsilon>0$ s.t. $B(f(a),\epsilon)\subset U$. Now the continuity of $f$ means we can find a $\delta>0$ s.t. $|a-x|<\delta$ implies $|f(a)-f(x)|<\epsilon$, which means $f(x)\in B(f(a),\epsilon)\subset U$. Thus $B(a,\delta)\subset f^{-1}(U)$ and thus $f^{-1}(U)$ is open.
Assume then that for every open $U$, $f^{-1}(U)$ is open. Let $a\in \mathbb{R}$ be arbitrary point and let $\epsilon>0$. Now $B(f(a),\epsilon)=:B$ is open and $a \in f^{-1}(B)$. By the assumption we can find a $\delta>0$ s.t. $B(a,\delta)\subset f^{-1}(B)$. This means that if $|x-a|<\delta$, then $|f(a)-f(x)|<\epsilon$, and thus $f$ is continuous at $a$. The continuity of $f$ follows since $a$ was arbitrary.
Now the equivalent claim for closed sets follows using Cheung SW's answer.
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Comments
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This is a question from my Real Analysis class and was hoping someone could answer it for me. The question from the beginning is:
Consider the function $f: \mathbb R \to\mathbb R$. Prove that $f$ is continuous iff for every closed set $K$ in $\mathbb R$, the set $f^{-1}(K)$ is closed in $\mathbb R$.
I keep running into dead ends because $K$ is closed but not bounded since its for every closed $K$ in $\mathbb R$.
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This depends on your definition of continuity. If you define it as "preimage of open is open," then the answer is straightforward. If you define it with epsilons and deltas, you may need to think a little harder. In either case, your definition will influence how we answer your question.
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I never knew the definition of continuity could be defined as such. I started off using the statement: For any epsilon (e = epsilon) > 0, there exists a delta (d = delta) such that x-d<Xo< x+d -> f(x)-e < f(Xo) < f(x)+e
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If I were to prove the question using epsilons and deltas, how would I do so?
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Regarding the only if part: $f(x)$ could also sneak in $\{f(x_n)\}_{n=1}^{\infty}$ if for several $x'\in X: f(x')=f(x)$. If one wants to be very precise, one should exclude the elements $n$ such that $f(x_n) = f(x)$
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Source: https://9to5science.com/prove-f-is-continuous-iff-for-every-closed-set-k-in-mathbb-r-the-set-f-1-k-is-closed-in-mathbb-r
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