This is a question from Section 3.2 of Wayne Patty's topology.  Question :Prove that if A is countable subset of $\mathbb{R}^2$ , then $\mathbb{R}^2 - A$ is pathwise connected.  The intution is very easy to get but I am not getting ideas on how should I define a continuous function f such that f(0)= a and f(1)=b for any a , b in $\mathbb{R}^2$. Kindly guide me with that.
38395
Here's a question that I share:  Show that if $D$ is a countable subset of $\mathbb R^2$ (provided with its usual topology) then $X=\mathbb R^2 \backslash D $ is arcwise connected.
42854
Please let me know if you find my proof satisfactory.  Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.  If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$. If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon &gt; 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.
38375
Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$  My first thought is that there can not be a continuous bijection, $f$, from $[0,1]$ to $\mathbb{R}$ because a continuous function that maps $[0,1]\rightarrow \mathbb{R}$ must be bounded so there can't be a surjection. So that would then be the proof.   Though I believe I am incorrect in my thinking because a hint for the assignment says to use the intermediate value theorem. That is "Suppose $f:[a,b]\rightarrow\mathbb{R}$ is continuous. If $f(a)&lt;\delta&lt;f(b)$ or $f(b)&lt;\delta&lt;f(a)$ then $\delta=f(c)$ for some $c\in[a,b]$".   Why is my first thought wrong and how is the IVT useful?
84302
I am looking for a set in the plane (with respect to the natural Euclidean topology) that is connected, locally connected, path-connected but not locally path-connected. I did not find one in , but maybe I overlooked?
73765
It'd be great if someone checked the proof I did for the following problem:  $f:X\longrightarrow Y$,   $f(\overline{A})\subseteq \overline{f(A)},\forall A\subseteq X \Leftrightarrow f$ continuous  proof:  Suppose $f(\overline{A})\subseteq \overline{f(A)}$ for any $A\subseteq X$ .  Let $C\subseteq Y$ be closed and define $A=f^{-1}(C)$.  $f(\overline{A})\subseteq \overline{f(A)}\subseteq\overline{C}=C$, therefore $\overline{A}\subseteq f^{-1}(C)=A$,  so $A$ is closed.  *Conversely, suppose $f$ is continuous.    Let $p\in\overline{A}$ and let $V$ be an open neighborhood of $f(p)$. Since $f$ is continuous,  $f^{-1}(V)$ is open. On the other hand $p\in f^{-1}(V)$, but since $p\in\overline{A}$ we have that $f^{-1}(V)$ contains points of  $A$. Therefore $V$ contains points of $f(A)$. Since $V$ is an arbitrary neighborhood of $f(p)$, we have $f(p)\in \overline{f(A)}$. So finally, $f(\overline{A})\subseteq\overline{f(A)}$.  Thank you very much!
52054
Consider the function $ f:\mathbb{R^2} \rightarrow \mathbb{R}$ given by   $ f(x,y) = \left\{   \begin{array}{l l}     \frac{xy}{x^2 + y^2} &amp; \quad \text{if (x,y) $\neq$ (0,0)}\\     0 &amp; \quad \text{if (x,y) = (0,0)}   \end{array} \right.$  Show that f is continuous along every horizontal and every vertical line (i.e. for every $x_0, y_0 \epsilon \mathbb{R}$, the functions $g,h:\mathbb{R}\rightarrow \mathbb{R}$ given by $g(t) = f(x_0,t)$ and $h(t)=f(t,y_0)$ are continuous).  I know that this function is not continuous at $(0,0)$, but the horizontal/vertical line issue is proving difficult to work with. Any help/hints appreciated!
81589
Is $(\mathbb{R},\tau_{co})$ compact where $\tau_{co}$ is the cocountable topology on $\mathbb{R}$?  I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says:     $\mathbb{N}-\{n\}$ is infinite countable for all $n\in\mathbb{N}$.   Then $V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}$ for all $n\in\mathbb{N}$.      Then $U=\{ V_n : n\in\mathbb{N}\}$ is a open covering of $(\mathbb{R},\tau_{co})$.      $U$ does not contain a countable subcover since $V_n\cap\mathbb{N}=\{n\}$ for each $n\in\mathbb{N}$ and $\mathbb{N}$ is infinite.   How would you solve that problem?
47071
Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
41991
Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a continuous real function and we consider its graph identified by the subset:   $$\mathcal{G}(f)=\{(x,f(x))\in\mathbb{R}^{n+1}:x\in\mathbb{R}^n\}$$  of $\mathbb{R}^{n+1}$ equipped with the usual euclidean topology that induced on $\mathcal{G}(f)$ a subspace topology. Well with these condictions we demonstrate that $\mathcal{G}(f)$ is homeomorphic to $\mathbb{R}^n$. So we consider the function   $$ h:\mathcal{G}(f)\owns(x,f(x))\rightarrow x\in \mathbb{R}^n$$  since the domain of $f$ is $\mathbb{R}^n$, clearly $h$ is surjective on $\mathbb{R}^n$ and then is also injective because if $h(x,f(x))=h(y,f(y))$ then $x=y$; now to prove the assertion we have to demonstrate of that $h$ and $h^{-1}$ are continuous funcion or that $h$ is open and continuous, but unfortunately I'm not be able to do this so can someone help me?
17718
Show that if $f: \mathbb{R} \to \mathbb{R}$ is a continuous injective map, then it is strictly monotonic.    Could someone give me a proof for this? I have the intuition for why it's true - I'm just having trouble expressing that intuition in a rigorous manner. Basically consider two points $x_1, x_2 \in \mathbb{R}$. By the problem statement, $f$ is continuous on $[x_1, \, x_2]$. WLOG, assume that $f$ is strictly increasing. It there exists a point where it is not increasing, then $f$ hits a value twice, and it's not injective.
61321
Let $T$ and $T^{'}$ be topologies on set $X$ such that $T \leq T^{'}$. If $(X,T^{'})$ is path-connected, is $(X,T)$ path-connected?  I think it is, and I tried to prove it by contradiction, but I got stuck. Any hint helps!
49805
As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
71197
1) Stuck with this problem, can you please help? Show that if $X$ is connected and for a connected subspace $A$ of $X$ we have $X \setminus A = U \cup V$ where $U,V$ are open in $X \setminus A$ and disjoint, then the sets $A \cup U$ and $A \cup V$ are connected.  2) The following result appears in Encyclopaedia of mathematics by Hazewinkel page $375$:  Any compact subset of the Sorgenfrey line is countable.  I tried to prove this without any luck. Why is this true?
20213
How do I show that $X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}$ is path connected?    Note that $X$ is a topological space with subspace topology $\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}$.  I only know two things (with respect to path connectedness):   A topological space $X$ is path connected iff for every $p,q \in X$, there is a continuous function $f: [a,b] \to X$ with $f(a)=p$ and $f(b) = q$.    Continuity $f$ is defined as for every open set in $X$, its preimage is open in $[a,b]$ (with subspace topology induced by $\mathbb{R}$ (with standard topology) )   Suppose $X$ and $Y$ are two topological spaces and $f: X \to Y$ is a continuous surjection. If $X$ is path connected, then $Y$ is path connected.
92802
My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!!
23925
If $ f: \mathbb{R} \to \mathbb{R} $ is continuous, and $\forall x \in \mathbb{R} :\;(f \circ f \circ f)(x) = x  $, show that $ f(x) = x $.  The condition that $f$ is continuous on $\mathbb{R}$ is crucial for the proof. I can find functions such as $\displaystyle\frac{x-3}{x+1}$ that satisfies  $ (f \circ f \circ f)(x) = x $. I have tried to negate the conclusion to see if there's a contradiction, but got stuck.
40025
I want to describe, up to homeomorphism, all the proper connected subsets of $\mathbb{R}$. I know the theorem that $A \subset \mathbb{R}$ is connected if and only if $A$ is an interval.  So consider the following intervals:  $[a,b), [a,b], (a,b)$  Note $[a,b)$ is not homeomorphic to $[a,b]$ (compactness argument) and $[a,b)$ is not homeomorphic to $(a,b)$ (by connectedness). Similary $[a,b]$ is not homeomorphic to $[a,b)$ (remove a,b from [a,b]).  Now any ray $(a,\infty)$ is homeomorphic to $(1,\infty)$ which is homeomorphic to $(0,1)$ and $(0,1)$ is homeomorphic to $(a,b)$. Similarly the ray $[-a,\infty)$ is homeomorphic to $(0,1]$. So I think this covers all the cases.   So I believe the answer is $3$, yes?