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Let $f:[0,1]\to\mathbb R^n$ such that $f(t)=ty+(1-t)x$ for some $x,y \in \mathbb R^n$. Prove that $f$ is continuous.

I know a definition that A function $f\colon X \rightarrow Y$ between two topological spaces $X$ and $Y$ is continuous if for every open set $V \subset Y$, the inverse image $f^{-1}(V) = \{x \in X \; | \; f(x) \in V \}$ is an open subset of $X.$

However I don't know how to apply the definition to solve the problem. Any help would be appreciated.

user
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    What else do you know about limits and continuity? Do you know "limit of the sums" theorem? Did you every do $\epsilon$-$\delta$-proofs in your calculus class? – Moishe Kohan Jan 26 '15 at 05:40
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    $\epsilon-\delta-$ proof is used for real-valued functions so I think it couldn't applied for my case. – user Jan 26 '15 at 05:45
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    If you understood these proofs for real-valued functions, you will easily extend the argument to the case of vector-valued functions. If not, what textbook are you using? Ddid you already cover continuity criterion for vector-valued functions in terms of their components? – Moishe Kohan Jan 26 '15 at 05:47
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    I don't think you can define the mapping by: f(t) = ty + (1 - t)x, "for all x and y in R^n". The mapping isn't well-defined. –  Jan 26 '15 at 06:19
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    Indeed. For all $x,y$, $f(1/2)=(x+y)/2$? What sense does that make? – Gerry Myerson Jan 26 '15 at 06:29
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    I think, user confused "for all" and "for some". – Moishe Kohan Jan 26 '15 at 06:45

2 Answers2

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If $x = y$, then $f(t) = x$ for all $t\in [0,1]$. Hence, for any open subset $V$ of $\Bbb R^n$, $f^{-1}(V)$ is either $[0,1]$ or $\emptyset$ (depending on whether $x\in V$ or $x\notin V$), both of which are open in $[0,1]$. So $f$ is continuous.

Now suppose $x\neq y$. Let $V$ be an open subset of $\Bbb R^n$. Given $t\in f^{-1}(V)$, there exists (by openness of $V$) an $\varepsilon > 0$ such that $B_\varepsilon(f(t)) \subseteq V$. Set $\delta = \frac{\varepsilon}{\|x - y\|}$. Then $f(B_\delta(t)\cap [0,1]) \subseteq B_\varepsilon(f(t))$. Indeed, given $u = f(s) \in f(B_\delta(t)\cap[0,1])$,

$$|f(t) - u| = \|(s - t)(x -y)\|= |s - t|\|x - y\| < \frac{\varepsilon}{\|x - y\|}\|x - y\| = \varepsilon.$$

Hence $B_\delta(t) \cap[0,1]\subseteq f^{-1}(B_\varepsilon(f(t))) \subseteq f^{-1}(V)$, showing that $f^{-1}(V)$ is open in $[0,1]$. Since $V$ was arbitrary, $f$ is continuous.

kobe
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  • For the case $x=y \in V$, why $[0,1]$ is open in $[0,1]$? – user May 28 '15 at 05:10
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    @user, $[0,1]$ is open in itself, regardless of whether $x = y$ or $x \neq y$. – kobe May 28 '15 at 05:12
  • Thank you. What is the topology in $[0,1]$? – user May 28 '15 at 06:59
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    @user here it's the subspace topology. Open sets in $[0,1]$ are of the form $A\cap [0,1]$ where $A$ is open in $\Bbb R$. – kobe May 28 '15 at 13:31
  • Thank you, I got it. By that definition, $[0,1]$ is also a closed set on itself, isn't it? I have one more question, is it true that $f^{-1}(\mathbb R^n) =[0,1]?$ for every function $f$? – user May 29 '15 at 00:56
  • And does your proof work for the Euclidean space instead of $\mathbb R^n$? – user May 29 '15 at 01:03
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    @user yes, $[0,1]$ is closed in itself. It's not true that $f^{-1}(\Bbb R^n) = [0,1]$ for every $f$. The question itself is not well posed, because there are many functions whose images are not contained in $\Bbb R^n$. I don't understand what you mean by the last question, Euclidean spaces are of the form $\Bbb R^k$. – kobe May 29 '15 at 03:28
  • By definition the inverse image $f^{-1}(V) = {x \in X ; | ; f(x) \in V }$, can we deduce that $f^{-1}(\mathbb R^n) = [0,1]$? For which type of $f$ that equality can be occurred? – user May 29 '15 at 03:35
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    @user by $f$ did you mean $f(t) = ty + (1 - t)x$? – kobe May 29 '15 at 03:42
  • $f(t)=ty+(1-t)x$ is the special case that I think $f^{-1}(\mathbb R^n)=[0,1]$ would hold. In general, do all continuous functions $f$ work? – user May 29 '15 at 03:50
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    @user ok, I understand your question now. The answer is still no. Take for instance $f : \Bbb R \to \Bbb R $ given by $f(x) = x$. It's continuous, but $f^{-1}(\Bbb R) = \Bbb R \neq [0,1]$. – kobe May 29 '15 at 03:59
  • Thanks. If I consider a function $f:X \to Y$, for which type of $f$ so that $f^{-1}(Y)=X$? – user May 29 '15 at 05:25
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    @user, given $f : X \to Y$, it's a fact that $f^{-1}(Y) = X$. For $f(x) \in Y$ for all $x \in X$, thus $x \in f^{-1}(Y)$ for all $x \in X$, i.e., $X \subset f^{-1}(Y)$. By definition of $f^{-1}(Y)$, $f^{-1}(Y) \subset X$. Thus $f^{-1}(Y) = X$. – kobe May 30 '15 at 11:55
  • Thank you. I got everything that I needed. – user May 30 '15 at 13:46
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It seems that you are looking at the function $$f:\quad[0,1]\to{\mathbb R}^n, \qquad t\mapsto (1-t)x+ty$$ for given and fixed $x$, $y\in{\mathbb R}^n$. Since for arbitrary $t$, $t'\in[0,1]$ one has $$f(t)-f(t')=(t-t')(y-x)$$ it follows that $$\bigl\|f(t)-f(t')\bigr\|\leq \|y-x\|\>|t-t'|\ .$$ This proves that $f$ is even Lipschitz continuous on $[0,1]$.

Christian Blatter
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