# $f:U\rightarrow \mathbb{R}$ continuous, injective- prove $f(U)$ is open.

If we have that $f:U\rightarrow \mathbb{R}$ where $U$ is open and $U\subset\mathbb{R}$ and $f$ is continuous and injective I want to show that $f(U)$ is open.

So I have considered the function restricted to $f:U\rightarrow f(U)$ so that it is a bijection.

I've had a few thoughts on this but not really gotten anywhere. I thought about just considering the map on an open balls so that the iamge had to be connected and then try to show that these are open but this didnt really get me anywhere.

I feel that I might be supposed to use the intermediate value theorem as this is the $1$ dimensional case of part of the invariance of dimension theorem for euclidean spaces, the proof of which relies on the Brower fixed point theorem.

I would prefer a hint as to how to do this, unless of course it is really simple and I am being very stupid.

Thanks very much for any help.

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Think of $f(x)=x$ and $U=[-1,1]$. –  Frank Science Jan 7 '13 at 11:20
I assume you want $U$ to be an open interval of $\Bbb R$, or at least an open subset of $\Bbb R$? –  Olivier Bégassat Jan 7 '13 at 11:21
I think f will be monotonic. –  Koushik Jan 7 '13 at 11:26
doesn't it follow from intermediate property –  Koushik Jan 7 '13 at 11:29
@OlivierBégassat Yeah thanks I've edited it now. –  hmmmm Jan 7 '13 at 11:37

It sufficies to show that for any $a,b\in \mathbb R$ with $a<b$, $f(a,b)$ is an open interval.

Since $f$ is continuous, $f[a,b]$ is compact and connected in $\mathbb R$, hence $f[a,b]=[c,d]$, for some $c,d\in \mathbb R$, let us see $f(x)\neq d$ for all $x\in (a,b)$. Suppose $d=f(x)$ for some $x\in (a,b)$, then since $f$ is injective $f(a)<d$ and $f(b)<d$, hence we can choose $r\in (c,d)$ such that $f(a)<r$ and $f(b)<r$. Then by the mean-value theorem there is $y\in (a,x)$ such that $f(y)=r$ and there exists $z\in(x,b)$ with $f(z)=r$, contradicting the injectivity of $f$. Thus $f(x)\neq d$ for all $x\in (a,b)$. Similarly $f(x)\neq c$ for all $x\in (a,b)$. Hence $f(a,b)=(c,d)$, since $f(a,b)$ is connected and $f[\{a,b\}]=\{c,d\}$.

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Theorems and proofs from Rudin's Principles of Mathematical Analysis:

4.14 Theorem Suppose $f$ is a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f(X)$ is compact.

Proof Let $\{V_\alpha\}$ be an open cover of $f(X)$. SInce $f$ is continuous, we have each of the sets $f^{-1}(V_\alpha)$ is open. Since $X$ is compact, there are finitely many indices, say $\alpha_1,\ldots,\alpha_n$, such that $$X\subset f^{-1}(V_{\alpha_1})\cup\cdots\cup f^{-1}(V_{\alpha_n})$$ Since $f(f^{-1}(E))\subset E$ for every $E\subset Y$, we have $$f(X)\subset V_{\alpha_1}\cup\cdots\cup V_{\alpha_n}$$ This completes the proof

4.17 Theorem Suppose $f$ is a continuous 1-1 mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f^{-1}$ defined on $Y$ by $$f^{-1}(f(x))=x\qquad(x\in X)$$ is a continuous mapping of $Y$ onto $X$.

Proof It suffices to prove that $f(V)$ is an open set in $Y$ for every open set $V$ in $X$. Fix such a set $V$.

The complement $V^c$ of $V$ is closed in $X$, hence compact; hence $f(V^c)$ is a compact subset of $Y$ and so is closed in $Y$. Since $f$ is one-to-one and onto, $f(V)$ is the complement of $f(V^c)$. Hence $f(V)$ is open.

Now we can claim the following proposition in a general setting:

Proposition Suppose $f$ is a continuous 1-1 mapping of locally compact metric space $X$ into metric space $Y$, and $V$ is an open subset of $X$. Then $f(V)$ is open in $Y$.

Hint For each $p\in V$, there's a (closed) neighborhood $N\subset V$ of $p$, such that $N$ is compact and $f(N)\subset f(X)$ is bounded in $Y$. Apply Theorem 4.17 into restriction map $f\big|_N$.

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A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing.

f must be monotonic by intermediate value theorem.so it maps an open set which must be a collection of open intervals to a open set.

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This is false. Consider $f : \mathbb{R} \setminus \{ 0 \} \to \mathbb{R}$ given by $f(x) = \frac{1}{x}$. This is continuous and injective but not monotonic. It is true when the domain of $f$ is a connected open subset of $\mathbb{R}$, but this is not a hypothesis here. –  Clive Newstead Jan 7 '13 at 11:35
@CliveNewstead yeah but can we not consider the basis of open balls which are connected and so if f maps an open ball to another open set then we just take the union of these. –  hmmmm Jan 7 '13 at 11:38
@hmmmm: What does the fact that the basis sets are connected have to do with it? The union of connected sets might not be connected. For example, $(0,1)$ and $(2,3)$ are connected but $(0,1) \cup (2,3)$ is disconnected. Your function must be monotonic on its connected components, but it needn't be monotonic on the whole domain $-$ between the connected components it can jump around and change direction all it likes! I mean, seeing is believing $-$ I've shown you a non-monotonic continuous injection! –  Clive Newstead Jan 7 '13 at 11:40
...but you can still use the result in K.Ghosh's answer by considering the restrictions of $f$ to the connected components of its domain separately. –  Clive Newstead Jan 7 '13 at 11:41
@CliveNewstead sorry what I meant was consider the function $f$ acting on the basis of open balls of $U$. Each open balls is connected and so then apply the result to each open ball. Just like you were saying with the connected components? –  hmmmm Jan 7 '13 at 11:44