Reputation
16,382
Top tag
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
2 19 49
Newest
 Enlightened
Impact
~139k people reached

4h
comment Graph Theory and vertices
An Euler trail has to use every edge of the graph once and only once. The triangle with 3 vertices would be a closed Euler trail if the 3 vertices and 3 edges of the triangle were THE WHOLE GRAPH. The graph we're talking about has 7 vertices and SEVEN EDGES. An Euler trail for that graph would have to use all 7 edges.
5h
comment Graph Theory and vertices
You lost me. Part (c) asks, does there exist a simple graph with 7 vertices and degrees 2,2,2,2,2,2,2 that contains no closed Euler trail. I showed you how to draw a simple graph with 7 vertices and degrees 2,2,2,2,2,2,2 that contains no closed Euler trail. What statement is wrong??
5h
comment Graph Theory and vertices
Sure, you can draw "7 vertices connecting each other", but that's not what problem (c) asks for. It asks you if you can draw a graph with 7 vertices, all of degree 2, which contains no closed Euler trail. And I just showed you how to draw one.
5h
comment Graph Theory and vertices
Your answer to (c) is wrong. Can you draw a simple graph with 3 vertices with degrees 2, 2, 2? OK, now can you draw a simple graph with 4 vertices with degrees 2, 2, 2, 2? OK, now put the two drawings side by side. Do you see 7 vertices with degrees 2, 2, 2, 2, 2, 2, 2? Can you draw a closed Euler trail?>
1d
comment Finite subset of $\Bbb R$ is nowhere dense
What is your definition of "nowhere dense"?
1d
comment Why imaginary numbers axis is plotted perpendicular to the real numbers axis?
Where else would you plot them?
1d
comment There exist a set $X$ such that the number of function $y: x\to \{1,2,3\}$ is $1000$.
I don't understand. Your question starts off "There exist a set $X$ such that . . ." and $X$ is never mentioned again.
2d
comment Equivalence relation for which there are infinitely many equivalence classes.
For instance, the real fourth roots of $\frac35$ will forn one equivalence class, and the real fourth roots of $0$ will form another equivalence class; however, the real fourth roots of $-1$ will not form an equivalence class.
May
23
comment A base generates an unique topology?
First, there is a definition of what it means for something to be a basis for a given topology. The viewpoint is that you have a topology to start with and you find a basis. For example, the Euclidean topology of the plane, the basis could be open discs, or open rectangles, or something else. What you quoted is the answer to the question: if I'm just given a set $X$ and a collection $\mathcal B$ of subsets, is this a basis for some topology? The answer is yes iff the stated conditions hold. The topology is unique; it consists of all unions of subcollections of $\mathcal B.$
May
23
comment A base generates an unique topology?
The quoted definition puts conditions on the basis $\mathcal B$ but does not mention a topology $\mathcal T.$ It tells you when $\mathcal B$ is a basis for a topology, but does not tell you what topology it's a basis for. So how did you pick $\mathcal T$ and $\mathcal T'$? Do think that $\mathcal B$ is a basis for the topology $\mathcal T''=\{\emptyset,X\}$? If not, why not?
May
23
comment Why a particular type of definition for increasing sequence is chosen?
@WillJagy: If the "vice-versa" means that $a_m\ge a_n$ implies $m\ge n,$ I'd say that makes a difference. The sequence $1,1,2,3,5,\dots$ is increasing according to the first definition. It's not increasing according to the second definition, because $a_1\ge a_2$ while $1\not\ge2.$
May
23
comment Why a particular type of definition for increasing sequence is chosen?
Grammatically, your second definition is missing one "if". Better: A sequence is increasing if it is the case that if $m\ge n$ then $a_m\ge a_n$ and vice versa.
May
22
comment Determine if $(((13)),\circ)$ is a normal subgroup of $(S_{3},\circ)$
What does $\Delta$ mean?
May
22
comment Little confusion about connectedness
Why such a complicated example?. Here's a simpler one. $[0,2]=[0,1)\cup[1,2].$ Now, $[0,1)$ and $[1,2]$ are both connected, and $[0,1)\cap[1,2]=\emptyset.$ Now I guess you think it follows that $[0,2]$ is disconnected. Why do you think that follows? What definition or theorem are you using?
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
$X\subseteq X$ is right, and from $X\subseteq X$ it follows that $2^A\subseteq2^A,$ and also that $f(a)\subseteq f(a)$ for each $a$ in $A$; but I don't see how it follows that $f(a)\subseteq2^A.$
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
If you assume (as one often does) that a set can't be an element of itself, then it's even simpler: the set $A$ itself is an element of $2^A$ but is not an element of $A,$ so $2^A\not\subseteq A.$
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
The set $R=\{a\in A:a\notin a\}$ is an element of $2^A$ (since it's a subset of $A$), but it's not an element of $A.$
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
Does $pink\not\subset\{\emptyset,pink\}$ mean that there is some $x$ such that $x\in pink$ and $x\notin\{\emptyset,pink\}$? What might that $x$ be?
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
Why do you think "is $2^A\subseteq A$" (from the title) is related to the subsequent questions? Anyway, the answer to your title is "no"; $2^A$ is never a subset of $A$.
May
22
comment If $A$ is a non-empty set and $2^A$ is the power set of $A$. Is $2^A \subseteq A$?
What is "$f(a)$ is the elements of $2^A$" supposed to mean? Don't you mean "$f(a)$ is an element of $2^A$"?