Two-point tests show that connectedness can be detected by continuous functions. We now turn this idea around: continuous maps cannot tear a connected space into two separated pieces. Continuous images and intermediate values form the central link between metric connectedness and real analysis. The intermediate value theorem becomes a consequence of connectedness of intervals, rather than a separate fact. The focus keyword continuous images and intermediate values will appear throughout this lesson because it explains why connectedness is preserved under continuous mappings. :::theorem[Continuous Image of a Connected Space] Let $(X,d_X)$ be a connected metric space and let $f:(X,d_X)\to(Y,d_Y)$ be continuous. Then $f(X)$ is connected as a subspace of $Y$. ::: :::proof Given that $X$ is connected and $f:X\to Y$ is continuous. To prove that $f(X)$ is connected. Suppose, if possible, that $f(X)$ is disconnected. Then there exists a continuous onto mapping $g:f(X)\to X_0$, where $X_0$ is the discrete two-point space. The composition $g\circ f:X\to X_0$ is continuous and onto. Therefore, by the two-point test, $X$ is disconnected. A contradiction. Hence $f(X)$ is connected. $\square$ ::: :::theorem[Intermediate Value Theorem] Let $f:[a,b]\to\mathbb{R}$ be continuous. If $y$ lies between $f(a)$ and $f(b)$, then there exists $x\in[a,b]$ such that $f(x)=y$. ::: :::proof Given that $f:[a,b]\to\mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$. To prove that there exists $x\in[a,b]$ such that $f(x)=y$. Since $[a,b]$ is an interval, $[a,b]$ is connected. Since $f$ is continuous, $f([a,b])$ is connected. A connected subset of $\mathbb{R}$ is an interval. Therefore $f([a,b])$ is an interval. Since $f(a)$ and $f(b)$ both belong to $f([a,b])$, every real number between them belongs to $f([a,b])$. Hence there exists $x\in[a,b]$ such that $f(x)=y$. $\square$ ::: :::definition[Intermediate Value Property] Let $(X,d)$ be a metric space and let $f:X\to\mathbb{R}$ be a function. Then $f$ is said to have the $intermediate value property$ if whenever $u,v\in f(X)$ and $w$ lies between $u$ and $v$, there exists $x\in X$ such that $f(x)=w$. ::: :::theorem[Intermediate Value Criterion for Connectedness] Let $(X,d)$ be a metric space. If every continuous function $f:X\to\mathbb{R}$ has the intermediate value property, then $X$ is connected. ::: :::proof Given that every continuous real-valued function on $X$ has the intermediate value property. To prove that $X$ is connected. Suppose, if possible, that $X$ is disconnected. Then there exists a continuous onto mapping $g:X\to X_0$. Define $h:X_0\to\mathbb{R}$ by $h(0)=0$ and $h(1)=1$. Since $X_0$ is discrete, $h$ is continuous. Therefore $h\circ g:X\to\mathbb{R}$ is continuous. The values $0$ and $1$ belong to $(h\circ g)(X)$, but $1/2$ does not belong to $(h\circ g)(X)$. Thus $h\circ g$ does not have the intermediate value property. A contradiction. Hence $X$ is connected. $\square$ ::: :::theorem[Fixed Point Theorem on a Closed Interval] Let $I=[-1,1]$ and let $f:I\to I$ be continuous. Then there exists $c\in I$ such that $f(c)=c$. ::: :::proof Given that $f:[-1,1]\to[-1,1]$ is continuous. To prove that $f$ has a fixed point. Define $g:[-1,1]\to\mathbb{R}$ by $g(x)=f(x)-x$. Then $g$ is continuous. Since $f(-1)\in[-1,1]$, we have $g(-1)=f(-1)+1\geq 0$. Since $f(1)\in[-1,1]$, we have $g(1)=f(1)-1\leq 0$. If either endpoint gives equality, then that endpoint is a fixed point. Otherwise $g(-1)>0$ and $g(1)<0$. By the intermediate value theorem, there exists $c\in[-1,1]$ such that $g(c)=0$. Therefore $f(c)-c=0$. Hence $f(c)=c$. $\square$ ::: :::example[No Fixed Point on a Half-Open Interval] Let $f:[-1,1)\to[-1,1)$ be defined by $f(t)=(t+1)/2$. If $f(t)=t$, then $(t+1)/2=t$, so $t=1$. But $1\notin[-1,1)$. Hence $f$ has no fixed point on $[-1,1)$. $\square$ ::: :::example[No Fixed Point on an Unbounded Interval] Let $f:[-1,\infty)\to[-1,\infty)$ be defined by $f(t)=t+1$. If $f(t)=t$, then $t+1=t$, which is impossible. Hence $f$ has no fixed point on $[-1,\infty)$. $\square$ ::: :::exercise [1] State the theorem on continuous images of connected spaces. [2] Prove the intermediate value theorem using connectedness. [3] Give an example showing that the fixed point theorem may fail on $[-1,1)$. [4] Explain why the image of a connected set under a continuous real-valued function is an interval. ::: :::answer [1] If $X$ is connected and $f:X\to Y$ is continuous, then $f(X)$ is connected. [2] Since $[a,b]$ is connected, its continuous image is connected, and connected subsets of $\mathbb{R}$ are intervals. [3] The map $f(t)=(t+1)/2$ has no fixed point in $[-1,1)$. [4] A connected subset of $\mathbb{R}$ is exactly an interval. ::: :::faq[Frequently Asked Questions] Q: Does a continuous image of a disconnected space have to be disconnected? A: No. A constant map sends any space to a singleton, which is connected. Q: Why does the intermediate value theorem follow from connectedness? A: Because continuous images of connected intervals are connected subsets of $\mathbb{R}$, hence intervals. Q: Why is the fixed point theorem stated on a closed interval? A: The endpoint behaviour and intermediate value argument may fail on half-open or unbounded intervals. ::: :::call-to-action[Continue to Closures and Enlargements] subtitle: Use continuous images and intermediate values before studying how connectedness behaves under closure. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/closures-and-enlargements-of-connected-sets button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/two-point-tests-for-connectedness :::