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As part of proofs on continuity, I should show that

(i) $\forall x \in \mathbb{R}^n,$ if $x_k \rightarrow x \implies f(x_k) \rightarrow f(x)$

implies

(ii) $f^{-1}(C)$ is closed for any closed set $C \in \mathbb{R}^n$

As usual, my approach is to write down clearly what this actually means.

For (i), that is:

$$ (1)\ \forall \epsilon' > 0\ \exists n_0' \in \mathbb{N}\ s.t. ||x - x_n|| < \epsilon' \ \forall n > n_0'$$ $$ \implies $$ $$ (2)\ \forall \epsilon'' > 0\ \exists n_0'' \in \mathbb{N}\ s.t. ||f(x) - f(x_n)|| < \epsilon'' \ \forall n > n_0''$$

For (ii):

$$ f^{-1}(C) = \{ x \in X: f(x) \in C\}\ closed\Leftrightarrow X \setminus \{ x \in X: f(x) \in C\}\ open = \{ x \in X: f(x) \in C^c\} = S$$

Now, for similar proofs in the past I went for a proof by contradiction, which I feel looks also promising here.

So I start by supposing that $S$ is not open, i.e. $ \exists y \in S$ s.t. $\forall r > 0 , B(y,r) \notin S$, i.e. $\exists a \in B(y,r)$ s.t. $a \notin S$.

Then I understand that I should continue that this somehow contradicts with $(1) \implies (2)$, but I can't quite imagine how to establish that.

Thus, I would be very grateful for any hints about how to continue from there.

Thanks in advance!

bonifaz
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2 Answers2

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Hint: Try to use the condition $\exists a \in B(y,r)$ s.t. $a \notin S$ to make a sequence $\{a_k\}$ that converges to $y \in S$. Note that $f(y_k) \in C$ for all $y_k$. Now apply (1). The fact that closed sets in $\mathbb{R}^n$ have the property that if $C$ is closed and $\{a_k\}$ is a sequence in $C$ that converges to $c$, then $c \in C$, will be handy.

Oliver E. Anderson
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The first condition means that $f$ is continuous. The second condition is a characterization of continuity of $f$. So from the first condition we have that $f$ is continuous. Thus if $x\in \Bbb R^{n}$ then for every neightbourhood $U$ of $f(x) $ there is an neightbourhood $V_{x}$ of $x$ such that $f(V_{x})\subset U$.

Thus if we take an open neightbourhood $U$ of $f(x)$ then $x\in f^{-1}(U)$. Because $f$ is continuous, there is a neightbourhood $V_{x}$ of $x$ such that $f(V_{x})\subset U$=>$V_{x}\subset f^{-1}(U)$=>$f^{-1}(U)$ is open.Now it's easy to show that for every $A$ closed in $\Bbb R^{n}$ $f^{-1}(A)$ is closed.

Haha
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