I know about Connectedness and Path Connectedness in Topology. Recently, I got stuck into a problem. Let $A$ be the following subset of $\mathbb{R}^2:$ $$A=\{(x,y) \mid (x+1)^2 + y^2 \leq 1\}\cup \{(x,y) \mid y=x\sin(\frac{1} {x}),\ x&gt;0\}$$ The connectedness is obvious from if $A$ is connected then $cl(A)$ is also connected. But I can't prove the path connectedness ( The answer is given that $A$ is both connected &amp; Path connected). Please help. Thanks in advance.
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Let $A$ be the following subset of $\mathbb{R}^{2}$:$$A=\{(x,y):(x+1)^{2}+y^{2}\leq 1\}\cup\{(x,y):y=x\sin(\frac{1}{x}),x&gt;0\}$$ Then  $1.$ $A$ is connected.  $2.$ $A$ is compact.  $3.$ $A$ is path connected.  $4.$ $A$ is bounded.  It is clear that $A$ is unbounded as the graph of $x\sin(\frac{1}{x})$, so option $2$nd and $4$th are not correct. Again $A$ is connected by using the fact that closure of connected set is connected. But i am confused about path connectedness of $A$. Please help me about path connectedness of $A.$ Thanking you.
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On $\mathbb R^2$ I consider the following set $G:=\{(x,\sin(\frac{1}{x}))| x\in (0,1]\}$   Well this set is pathwise connected (as a graph of a contionous function on an interval). The function is also oscillating on the left hand side so it makes sense to write the closure as $\bar{G}=G\cup(\{0\}\times[-1,1])$ but how can this be proved formally?  The next thing which is not clear to me is why $\bar{G}$ is connected but not pathwise connected, so there is no contin. way $c:[0,1]\rightarrow \bar{G}$ with starting point $c(0)=x$ and $c(1)=y$
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The graph of $xy = 1$ in $\Bbb C^2$ is connected. True or false?  I know that it is not connected in $\Bbb R^2$, but what is the case of $\Bbb C^2$?
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I have two questions which pertain to differentiability, connectivity and path connectivity. Ocasionally, I will encounter an author who defines connectivity in the following way:   An open subset $U$ of $\mathbb{R}^n$ is said to be connected if and only if given two points $a$ and $b$ of $U$ there exists a differentiable mapping $\phi: \mathbb{R} \rightarrow U$ such that $\phi(0) = a$ and $\phi(1) = b$.   This particular example is from Edward's Advanced Calculus of Several Variables p 84. Now, this is obviously not the standard definition we learn from topology which has nothing to do with differentiability but rather whether there exists two nonempty open subsets that comprise a separation. It also seems to me that what the author is really defining what it means for a space to be "smoothly path connected", which of course implies connectivity and, it seems to me, considerably more.  My first question is: Is "smoothly path connected", as defined above actually equivalent to "connected", in the topological sense, in $\mathbb{R}^n$?  Next, in Vector Calculus by Baxandall and Liebeck on p 150 the authors state the existence of a continuous path $\alpha$ from the closed interval $[0,1]$ to an open subset $D$ of $\mathbb{R}^n$ where $\alpha (0) = a \in D$ and $\alpha(1) = b \in D$ guarantees the existence of a differentiable path with the same properties. This claim is stated without proof.   My second question is: Can someone provide a reference to a proof of the above claim or explain why it is so?
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Let for an open connected subset $U\subset \mathbb R^n$ and a number $\varepsilon &gt;0$: $$ U_\varepsilon=\{x\in U: dist (x, \partial U)&gt; \varepsilon \}. $$ Then  $U_\varepsilon$ is open but in general not connected.  But I suppose that for sufficiently small $\varepsilon &gt;0$ also  $U_\varepsilon$ is connected. How to prove its?  Edit. Let's assume that additionally $U$ is bounded. Is then $U_\varepsilon$ connected?
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If $U$ is a connected open set in $\mathbb R^n$, is $\bar U$ path-connected? (that is, for any two points $x_1,x_2$ in the closure of $U$, can we find a continuous path connecting them?)
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Let $A$ be a connected set in $\mathbb{S}^1$. Show that $A^c$ is also connected.  My idea: suppose $A^c$ is not connected. Then, there exist $U,V \subset \mathbb{S}^1$ open and disjoint such that $U \cup V =A^c  $. Therefore $\mathbb{S}^1 = A \cup U \cup V$. If I had that $A\cup U$ was open I could arrive at  contradiction, since $\mathbb{S}^1$ is connected, but I can't say that...
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If $\{E_\alpha\}_{\alpha\in A}$ is connected in $\mathbb{R}^n$, $\bigcap\limits_{\alpha\in A}E_\alpha \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E_\alpha$ is connected.  I have zero intuition on how to do a proof of this statement. Please help.  I started it by contradiction like this: suppose that $\bigcup\limits_{\alpha\in A}E_\alpha$ is not connected. Then there exist non-empty disjoint separating sets $U$ and $V$, such that $U$ and $V$ are relatively open in $\bigcup\limits_{\alpha\in A}E_\alpha$, and $U\cup V= \bigcup\limits_{\alpha\in A}E_\alpha$. But I have no idea what to do next.
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If I want to prove that $\mathbb{R} \setminus \mathbb{Q}$ is disconnected, does it suffice to say that there are two open disjoint sets that cover $\mathbb{R}\setminus\mathbb{Q}$, namely:  $$(- \infty, 0), (0, \infty)\text{ ?}$$  Along the same lines, I want to prove or disprove that $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$ (points $(x, y) \in S$ that have at least one irrationais connected. I feel that it is also disconnected; does it suffice to say that there are two open disjoint sets that cover $S = \mathbb{R}^2 \setminus \mathbb{Q}^2$, namely $((- \infty, - \infty), (0, 0))$ and $((0, 0), (\infty, \infty))$? (Very iffy on my assertion and my notation, sorry.)  Thanks!
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Let $V$ be a subspace of $\mathbb R^3$ generated by the vectors $\{(1,1,1),(0,1,1)\}.$ There are $3$ vectors $u_1=(0,0,1),u_2=(1,1,0),u_3=(1,0,1)$.Then which of the following sets are connected?  $1.\left(\mathbb R^3\setminus V\right)\cup\{(0,0,0)\}$  $2.\left(\mathbb R^3\setminus V\right)\cup \{tu_1+(1-t)u_3:t\in [0,1]\}$  $3.\left(\mathbb R^3\setminus V\right)\cup\{tu_1+(1-t)u_2:t\in [0,1]\}.$  $4.\left(\mathbb R^3\setminus V\right)\cup \{(t,2t,2t):t\in \mathbb R\}$  Now $V$ is generated by two linearly independent vectors so is like $\mathbb R^2.$ So $\left(\mathbb R^3\setminus V\right)$ will be disconnected but giving the $(0,0,0)$ with them will make it connected.  No. $4$, I think will be connected also because $\{(t,2t,2t),t\in \mathbb R\}$ is in $V$ and contains the point $(0,0,0)$ Now, any neighbourhood of this point will have points from both $V$ and $\left(\mathbb R^3\setminus V\right)$ and so the union is connected.I'm not sure about no. $2$ and $3.$
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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.
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I have an exercise found on a list but I didn't know how to proceed. Please, any tips?  Let $X$ be a connected subset of a connected metric space $M$. Show that for each connected component $C$ of $M\setminus X$ that $M\setminus C$ is connected.
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$A$ and $B$ are  two connected subset of $\mathbb{R}$ define $A+B=\{x+y:x\in A,y\in B\}$, then is $A+B$ also connected?  naturally I was thinking about two disjoint connected subsets of $\mathbb{R}$, say $A=[0,1]$ and $B=[4,5]$    then $A+B=[4,6]$? so is it ingeneral true?
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Prove that an open interval $(a,b)$ and a closed interval $[c,d]$ are not homeomorphic.  I'm trying to prove this statement but have only vague ideas on how to start. How may I use the property of connectedness to show this?
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It's well known that, in $\mathbb{R}^n$:     (1) Open and Connected $\Rightarrow$ Path-connected   The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.  Well, trying to see where (1) would keep being valid, I arrived at the following:     If $X$ is a locally convex topological vector space, then (1) is valid.   But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).  Anyway, given the previous considerations, my questions are:  Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?  And bonus question, since (2) implies (1) if the set is open.:  Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
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Is the proof of the following exercise correct?     Let $(X,\tau)$ be a connected topological space, $Y\subset X$ connected and $U\subset X\setminus Y$ clopen. Prove that $ Y \cup U$ is connected.   Proof  As $Y$ is connected, $X\setminus Y$ is still connected.  As $U\subset X\setminus Y$ is clopen and $X$ is connected (there is only one clopen set, namely $X$), we must have this $U=X$.  And finally $Y\cup U=Y\cup X=X$ is connected.
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Consider the following two definitions:  Connected : A topologiocal space X is connected if it is not the disjoint union of two open subsets, i.e. if X is a disjoint union of two open sets A and B, then A or B is empty set.  Path Connected : A topological space X is path-connected if any two points in X can be joined by a continuous path.    So far I can picture, I think they should be equivalent. If X is connected iff it is path-connected. But, the Thm I got from lecture is just if X path connected then X is connected (The converse not necessarily true.)  Can anyone give me some picture of space which is connected but not path-connected? Just picture please, cuz I wanna grab the idea. I have already seen some examples like in , I cannot grab the idea if it (such space) consist of a single piece, then I couldn't make a continuous path from any two points.  Thank you :)