I was reading Allen Hatcher's proof of the Brouwer's fixed point theorem using the no retraction theorem. That is, we want to prove that every continuous $f: \mathbb{D}^n \to \mathbb{D}^n$ has a fixed point $f(x) = x$, by contradiction.  The no retraction theorem says there is no continuous $r: \mathbb{S}^{n-1} \to \mathbb{D}^n$ such that $r \mid_{\mathbb{S}^{n-1}}(y) = y$. i.e. $r$ restricted to $\mathbb{S}^{n-1}$ is the identity functon.  Suppose a continuous $f$ exists with no fixed point. Then, define $r(x)$ by the point on the line $tx + (1-t)f(x)$ intersecting $\mathbb{S}^{n-1}$. The book, and lecture notes that I looked up, say this is clearly continuous.  Question: How do we proof that $r(x)$ is continuous?
123866
Take a continuous mapping $f: \bar{B^{n}} \rightarrow \bar{B^{n}}$, where $\bar{B^{n}}$ is a closed unit ball in $\mathbb{R}^{n}$. Assume that $f(x) \neq x$ for every $x \in \bar{B^{n}}$. Define another function $r$ by following the directed line segment from $f(x)$ through $x$ to its intersection with $\partial B^{n}$, and let the intersection point be $r(x)$. Is it immediately evident that $r$ is a continuous function? Is so how does it follow? Thanks.
12817
Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof:  $$ |d(x,a) - d(y,a)| \leq d(x,y) $$ for any $a\in A$ - but we cannot just replace it by $|d(x,A) - d(y,A)|\leq d(x,y)$ since the minimum (or infimum in general) can be attained in different points $a\in A$ for $x$ and $y$, so we only have that $$ |d(x,A)-d(y,A)|\leq d(x,y)+\sup\limits_{a,b\in A}d(a,b) $$ which does not mean continuity.
44734
Assume that $f: \mathbb{R} \rightarrow \mathbb{R}$  is differentiable.  If there exists an $L&lt;1$ such that for each $x \in \mathbb{R}$, $f'(x)&lt;L$, prove that there exists a unique point $x$ such that $f(x)=x$.  So far I have that where I fixed $a$ to be zero and came up with three cases to show existence for  $f(0) = 0$, $f(0) &lt; 0$, and $f(0) &gt; 0$  $f(0)=0$ was fairly trivial following from the mean value theorem but the other two cases I was only able to eliminate half of the possible fixed points by contradiction and I need help showing it explicitly.  Edit as to avoid confusion here is the work I have garnished so far:  From the mean value theorem it follows for some arbitrary $a,b \in \mathbb{R}$ and $a\neq b$ that there exists a point $c$ such that $a&lt;c&lt;b$ or $b&lt;c&lt;a$ then implies that $\frac{f(b)-f(a)}{b-a}=f'(c)$ and this is because it is differentiable on $\mathbb{R}$ and, by convention, is also continuous on $\mathbb{R}$ as well.  Then the result of $\dfrac{f(b)-f(a)}{b-a}=f'(c)$ then follows for $L&lt;1$ such that $\forall x \in \mathbb{R}$ means $f'(x)&lt;1$ from our hypothesis. Which then follows  (*) $\dfrac{f(b)-f(a)}{b-a}=f'(c)&lt;L&lt;1 \Rightarrow f(b)-f(a)&lt;L(b-a)&lt;b-a$  Now without loss of generality choose $a=0$ and let $b$ be some arbitrary value in $\mathbb{R}$ then there are three cases that need to be considered for the value of $f(a)$:  (i) $f(0) = 0$, (ii) $f(0) &lt; 0$, (iii) $f(0) &gt; 0$  (i) For $f(0)=0$   Then $0$ is a fixed point of point of $f(x)$ and it follows from (*) that if $f(0)=a=0$ then $f(b)-f(0)=f(b)&lt;Lb&lt;b$ which shows that $f(b)&lt;b$ thus shows that for any arbitrary value $b\in \mathbb{R}$ that no $f(b)=b$ thus shows that $0$ is the only fixed point if $f(0)=0$  (ii) For $f(0)&lt;0$  Suppose there exists a fixed point $d&gt;0$ then:  $\frac{f(d)-f(0)}{d}&lt;L&lt;1 \Rightarrow d-f(0)&lt; dL&lt;d \Rightarrow -f(0)&lt;d(L-1)&lt;0\Rightarrow 0&lt;-d(L-1)&lt;f(0)$ but this contradicts that $f(0)&lt;0$  thus no other fixed point exists if $d&gt;0$ when $f(0)&lt;0$  (iii) For $f(0)&gt;0$
79821
I tried looking up a question regarding graphs of continuous functions on this site, but all the ones I found consider functions from $\mathbb{R}$ into $\mathbb{R}$. I have been pondering the following question: given a general topological spaces $X, Y$, and a function $f: X\to Y$, when does $Graph(f)$ closed in $X\times Y$ imply that $f$ is continuous. By the closed graph theorem, this is true whenever $X$ and $Y$ are both Banach spaces.   Also, it is fairly easy to prove that whenever $Y$ is a Hausdorff space and $f$ is continuous, then $Graph(f)$ is closed, but I do not think that the converse is true, so I am trying to find an example where $X$ is some topological space, $Y$ a Hausdorff space, $f: X\to Y$ a function with a closed graph in $X\times Y$, but who fails to be continuous. As of yet I have not been able to find such a counterexample, partially because I have no clue where to look for such a counterexample. I would really appreciate getting some directions to go in.
105074
Here's Theorem 4.17 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:      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 $$ f^{-1}( f(x) ) = x \ \ \ \ \ \ \ ( x \in X ) $$ is a continuous mapping of $Y$ onto $X$.    Now in Example 4.21 Rudin illustrates that compactness of $X$ is essential in Theorem 4.17.   So here is Example 4.21:   Let $X$ be the half open interval $[0, 2 \pi )$ on the real line, and let $\mathbf{f}$ be the mapping of $X$ onto the circle $Y$ consisting of all points whose distance from the origin is $1$, given by $$ \mathbf{f}(t) = ( \cos t, \sin t ) \ \ \ \ \ \ ( 0 \leq t &lt; 2 \pi). $$ Then this function $\mathbf{f}$ is of course continuous and injective. But Rudin asserts that the inverse function is discontinuous at the point $(1, 0) = \mathbf{f}(0)$. How to demonstrate this fact rigorously?  So far in the book, Rudin has not discussed the functions $\cos$ and $\sin$. So can we come up with an example that uses only what Rudin has discussed?
58232
I need to prove the following:  There's no continuous function $f:[a,b]\to \mathbb{R}$ that takes each of its values $f(x)$, $x\in [a,b]$  exactly twice.  First of all, I didn't understand the question. For example $x^2$ takes $1$ twice, in the interval $[-1,1]$. Is it saying that it does not occur for all $x$ in the interval? But what about $f(x) = c$? Is it saying that it does not occur only exactly $2$ times, then? I have no idea about how to prove it. I know that for $f(x)$ such that $f(a)&lt;f(x)&lt;f(b)$, if $f$ is continuous then there is a $c\in [a,b]$ such that $f(c) = f(x)$.  Now, there's the following proof in my book and I really wanted to understand it, instead of just getting a new proof  Since the interval $[a,b]$ has only $2$ extreme points, then the maximum or minimum of $f$ must be in a point $c\in int([a,b])$ and and in another point $d\in [a,b]$. Then, there exists $\delta&gt;0$ such that in the intervals $[c-\delta, c), (c,c+\delta)$ (and if $d$ is not extreme of $[a,b]$, $[d-\delta, d]$) the function takes values that are less than $f(c) = f(d)$. Let $A$ be the greatest of the numbers $f(c-\delta), f(c+\delta), f(d-\delta)$. By the intermediate value theorem, there are $x\in [c-\delta, c), y\in (c, c+\delta]$ and $z\in [d-\delta, d)$ such that $f(x)=f(y)=f(z)=A$. Contradiction.   Well, why the last part? Why is it that I can apply the intermediate value theorem to these values? For example, $&lt;f(c-\delta)&lt;p&lt;f(c)$, then by the theorem I know that there exists $m\in [c-\delta, c)$ such that $f(m) = p$. Same for the other intervals. But what guarantees thhat the greatest of the values between  $x\in [c-\delta, c), y\in (c, c+\delta]$ will be inside the intervals $[c-\delta), c), (c,c+\delta), [d-\delta, d)$?
59414
How would you show that there is no continuous injective map  $f:\mathbb{R}^3 \to \mathbb{R}$?  I tried the approach where if $a \in \mathbb{R}$ then $\mathbb{R} \backslash \{a\} $ is not clearly connected so there can't exist a continuous function otherwise $\mathbb{R} \backslash \{a\} $ would be connected?
23490
If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.   Is the proof trivial as it seems or does it need an analytical approach?
39581
Consider a closed disc $D^n\subseteq \mathbb{R}^n$ and let $f:D^n\rightarrow \mathbb{R}^n$ be a continuous map with $f|_{S^{n-1}}$ is the identity on the boundary of $D^n$. Prove that $D^n \subseteq \operatorname{Im}(f)$. Is my following argument correct? Suppose $x\in D^n$ and $x\not \in f(D^n)$. Then we have $\operatorname{id}_{S^{n-1}}$ factors through $f\circ i:S^{n-1} \hookrightarrow D^n \rightarrow \mathbb{R}^n-\{x\}$. Recalling the fact that $\mathbb{R}^n-\{x\}$ deform retracts to $S^{n-1}$, the contradiction here comes from applying the $H_{n-1}$ functor to obtain that $H_{n-1}(f\circ i):\mathbb{Z}\rightarrow 0 \rightarrow \mathbb{Z}$ is the zero homomorphism, certainly not equal to the identity homomorphism $H_{n-1}(\operatorname{id}_{S^{n-1}}):\mathbb{Z}\rightarrow \mathbb{Z}$. Hence $f$ must contain its image. Also is there a way to prove this directly? I think I can smell Brouwer's fixed point theorem and/or Borsuk-Ulam but I'm not sure how to employ it.
46346
Brouwer's fixed point theorem:     Every continuous function $f$ from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point.   I am wondering why the word "convex" is in there.  It seems to me that it is necessary and sufficient for $K$ to have no holes, which is a weaker condition than convexity.  Necessary: if $K$ has a hole, then the continuous mapping that simply rotates points around this hole has no fixed point.  Sufficient: A unit disk in any number of dimensions has a Brouwer fixed point.  Any compact, hole-less set $K$ is homeomorphic to the unit disk in some number of dimensions.  If we map $K$ to the unit disk $D$ with a homeomorphism $h$, then consider the function from $D$ to $D$ given by $h \circ f \circ h^{-1}$.  This is a continuous function from the unit disk to itself (which is convex and compact), so it has a Brouwer fixed point $x$.  Then $h(f(h^{-1}(x))) = x$, so $f(h^{-1}(x)) = h^{-1}(x)$, so $h^{-1}(x)$ is a Brouwer fixed point of $K$.  What's wrong with this argument?
111097
Assume that $f_n → f$ uniformly on $S$ and each $f_n$ is continuous on $S$. Let $(x_n)$ be a sequence of points in $S$ converging to $x \in S$. Then $f_n(x_n) → f(x)$.  I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.
118717
I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.  Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that    for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function; if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.   The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then   $$\sum_{k=0}^{n-1} g(k/n) = 0$$  and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.  Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell &lt; 1/2$.  (For $\ell &gt; 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)  Can anyone help me out?  Cheers,
26179
How to prove that a continuous function, homotopic to the constant map $f:S^1\to S^1$  (a) has a fixed point and that (b) has a point $x$, such that $f$ maps $x$ to its antipodal point $-x$?
2881
Let $S = [0,1) \cup [2,3]$ and $f\colon S \rightarrow \mathbb R$ be such that $f(S)$ is connected . Which of the following are true:  a) $f$ is discontinuous exactly at one point.  b) $f$ is discontinuous exactly at two points.  c) $f$ is discontinuous at infinite many points.  d) $f$ is continous.  I think $f$ is continuous . We need to show that $\lim_{x \to a^+}f(x) = f(a)$ for any $a \in S$. Since $f(x)$ is increasing, $\lim_{x \to a^+}f(x) \leq f(a)$. Suppose $\lim_{x \to a^+}f(x) &lt; f(a)$. Then there is $l$ such that $\lim_{x \to a^+}f(x) &lt; l &lt;f(a)$, which contradicts connectedness of $f(S)$. Similarly we can show that $\lim_{x \to a^-}f(x) = f(a)$.  I would be thankful if anyone would check my solution and correct me if there are any errors.
67821
Assume  $f : I \rightarrow \mathbb{R}$ is a non-decreasing on an open interval $I$ and that $f$ satisfies the Intermediate value property or  on $I$ (that is, for any $a &lt; b$ in $I$ and any $L$ between $f(a)$ and $f(b)$, there exists $c \in [a, b]$ such that $ f (c) = L)$. Then, prove that $f$ is continuous.  However, I know that a function can be discontinuous and also satisfy the IVT at the same time. Could someone point me in the right direction?
95875
I got started recently on proofs about continuity and so on. So to start working with this on $n$-spaces I've selected to prove that every linear function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous at every $a \in \mathbb{R}^n$. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows:  Since $f$ is linear, we know that there's some $k\in \mathbb{R}$ such that $|f(x)|\leq k|x|$ for every $x\in \mathbb{R}^n$, in that case let $a\in \mathbb{R}^n$ and let $\varepsilon &gt;0$. Consider $\delta = \varepsilon /k$ and suppose $|x-a|&lt;\delta$, in that case we have:  $$|f(x)-f(a)|=|f(x-a)|\leq k |x-a|&lt;k \frac{\varepsilon}{k}=\varepsilon$$  And since $|x-a|&lt;\delta$ implies $|f(x)-f(a)|&lt;\varepsilon$ we have that $f$ is continuous at $a \in \mathbb{R}^n$. Since $a$ was arbitrary, $f$ is continous in $\mathbb{R}^n$. Is this proof fine? Or there was something I've missed on the way?
101557
Let, $id:(X,d_1)\to (X,d_2)$ is continuous. Then which is(/are) TRUE ?  (A) $X$ must be singleton.  (B) $X$ can be any finite set.  (C) $X$ can NOT be infinite  (D) $X$ may be infinite but NOT uncountable.  My Thought :  If $X$ is singleton then the function is continuous trivially but the condition is not necessary. So (A) is FALSE. If $X$ is finite then $d_1$ and $d_2$ are equivalent and so the function is continuous. So, (B) is TRUE. If $X=\mathbb R$ and $d_1$ is discrete metric and $d_2$ is any metric then then the function is continuous. So (C) and (D) are FALSE.  Am I right or wrong ? If I am wrong then where my fallacy ?