# Let $f(x)$ be an increasing function. Assume its image $f(C)$ is also a connected set. Prove that $f$ must be continuous

Let $f: R → R$ be an increasing function. Assume that for every connected subset $A$ of $R$, its image $f(A)$ is also a connected set. Prove that $f$ must be continuous.

To prove this, I am thinking it will be best to prove the contrapositive.

Assume $f$ is discontinuous. Then since $f$ is increasing, every discontinuity is a jump discontinuity, which would then disconnect the set $f(A)$. Since this is true, it must be true that the original statement is true.

I am having trouble writing up a formal proof, but I think my idea is correct

Your idea is pretty sound, except that you have not defined a particular set $A$ in your proposed solution before you refer to it. It is possible to flesh it out so that it works.

Or, you can quickly reduce this exercise to a well-known problem, whose solution you can google if you have trouble writing it up yourself.

The assumption that $f$ "preserves connectedness" implies that $f$ has the intermediate value property on $\mathbb{R}$:

Intermediate Value Property: A function $f \colon \mathbb{R} \rightarrow \mathbb{R}$ has the IVP if whenever $a < b$ and $y$ is a point between $f(a)$ and $f(b)$, there exists a point $c$ between $a$ and $b$ with $f(c) = y$.

So, show that preserving connectedness implies that $f$ has the IVP. It is then a well-known result that an increasing (or monotone) function with the IVP must be continuous.

• is assuming A is a subset of R not enough? This assumption was given in the question Apr 17, 2016 at 2:10
• The question as written doesn't assume that A is a particular subset of $\mathbb{R}$. The "for all" indicates that all subsets must be considered as possible $A$s. But your argument discusses a particular such set, and you'd need to pick out a particular one before you can discuss it. You can, for instance, pick a discontinuity $c$ of $f$ and then pick $A$ to be a connected set (i.e., interval, since you are working in $\mathbb{R}$) that contains $c$. Apr 17, 2016 at 2:14
• So, consider the discontinuity c of f. Let A be a connected interval [a,b] which contains c. Then f(A) is a disconnected set [a,c) U (c,b]. Is this correct? Where should I go from here? Apr 17, 2016 at 2:27
• $[a,c)$ and $(c,b]$ would both be $A$, not in $f(A)$. But using the definition of jump discontinuity, you can say there are numbers $w < z$ with $\lim_{x \to c^{-}} f(x) = w$ and $\lim_{x \to c^+} f(x) = z$. Further, you can argue that $f(a) < w < z < f(b)$. Argue that the image of $[a,b]$ under $f$ is the union of disjoint sets -- one contained in $[f(a),w)$ and the other contained in $(z,f(b)]$ -- such that neither contains a point in the closure of the other. The image of $[a,b]$ is therefore disconnected. Apr 17, 2016 at 2:32

Let $x\in R$. You have to show that for every $c>0$ there exists $d>0$ such that $\mid x-y\mid <d$ implies that $\mid f(x)-f(y)\mid c$. Suppose it is not true, and consider $c_n={1\over n}$ there exists $c_0\geq 0$ and $x_n$ such that $\mid x-x_n\mid< {1\over n}$ and $\mid f(x)-f(x_n)\mid\geq c_0$. There exists an infinite number of $x_n>x$ or $x_n<x$. Without restricting the generality, we suppose that there exists an increasing function $g:N\rightarrow N$ such that $y_n=x_{g(n)}$ satisfies $\mid y_n-y\mid <{1\over n}$, $\mid f(x)-f(y_n)\mid\geq c_0, y_n>x$. Since $f$ sends an interval to an intervall and $f$ is an increasing function, $f([y_n,\infty)\subset [f(y_n),\infty]\subset [f(x)+c_0,+\infty]$ it implies that for every $y>x$, $f(y)>f(x)+c_0$, thus $f([x,x+1])$ is not an interval, since it contains $f(x)$ but not the interval $[f(x)+f(x)+{c_0\over 2}]$ andt it is not reduced to a point. Contradiction.