Ff Continuous Preimage of Closed Set

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Preimage of closed set is closed implies function is continuous

  • Thread starter a6623
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a6623 Asks: Preimage of closed set is closed implies function is continuous
Let function $f$ be defined from $D \to \mathbb{R}$, $D \subset R$ closed. If the preimage of every closed set $A$ is closed, then $f$ is continuous.

I was trying to use the standard definition of continuity, so take arbitrary $x_0 \in D$ and show that $\forall \frac 1 m, \exists \frac 1 n. |x - x_0| < \frac 1 n \implies |f(x) - f(x_0) | < \frac 1 m$.

Take arbitrary $\frac 1 m$. Let $A$ be the closed set ranging from $[f(x_0 - \frac 1 m), f(x_0 + \frac 1 m)]$ and hence the closed preimage $f^{-1}(A)$ contains $x_0$ as well and also ranges over $[x_0 - \frac 1 m, x_0 + \frac 1 m]$.

Then we know that any $x \in f^{-1}(A)$ satisfies $|x - x_0| < \frac 1 m$ and it also follows that $f(x) \in A$ by definition of preimage, so $|f(x) - f(x_0)| < \frac 1 m$, showing convergence.

Does this proof work?

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enter image description here

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SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.

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SolveForum.com may not be responsible for the answers or solutions given to any question asked by the users. All Answers or responses are user generated answers and we do not have proof of its validity or correctness. Please vote for the answer that helped you in order to help others find out which is the most helpful answer. Questions labeled as solved may be solved or may not be solved depending on the type of question and the date posted for some posts may be scheduled to be deleted periodically. Do not hesitate to share your thoughts here to help others.

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