The open interval $(-\dfrac{\pi}{2}, \dfrac{\pi}{2})$ is well known to be homeomorphic to $\mathbb{R}$ through the homemorphism $f(x) = \tan(x)$  Is the closed interval also homeomorphic to the real line?   It seems to be the case until you hits the end points, which is undefined by $f(x) = \tan(x)$ so that seems to ruin everything.  Also, is the closed interval homeomorphic to open interval given any intervals?
48594
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?
70589
What is the cardinality of the set S of all closed intervals on the real number line with rational (positive) lengths?  I believe the number of intervals with a specific fixed length but varying start points is |R|, and the number of intervals with a fixed start point but varying lengths is |Q| = |N|. I think the answer to the question would then be |S| = |R x N| = |R|? I'm not quite sure how to prove the last equality though. Would I have to define a bijective function that maps elements of (R, N) to R? How would I approach doing so?
103077
Sierpiński's theorem states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found  and a discussion ).   An immediate corollary is that the rational intervals $[0,1],(0,1),[0,1)$ are homeomorphic. However, the theorem's proof is not very constructive and in fact quite complicated. (It's not a trivial theorem).   Is there a way to construct an explicit homeomorphism between these intervals?
80961
I don't understand the following proof of that every interval is connected in $\mathbb{R}$.     Let $Y$ be an interval in $\mathbb{R}$ and suppose that $Y$ is not connected.      Then $Y=A\cup B$, where $A,B\subseteq Y$ are open in $Y$, $A,B\neq\emptyset$ and $A\cap B=\emptyset$. Let $a\in A$ and $b\in B$. Without loss of generality, $a&lt;b$.       Let $\alpha=\sup\{x\in\mathbb{R}:[a,x)\cap Y\subseteq A\}$.      Then $\alpha\le b$ and $\alpha\in Y$. It is clear that $\alpha\in Cl_{Y}(A)$, and $A$ is closed in $Y$, then $\alpha\in A$. Since $A$ is open in $Y$, $Y$ is an interval, $b\in Y\setminus A$ and $\alpha &lt;b$, then there exists $r&gt;0$ such that $(\alpha-r,\alpha+r)\cap Y\subseteq A$. We can conclude that $[a,\alpha+r)\cap Y\subseteq A$, which is a contradiction.   I don't understand why $\alpha\le b$ and $\alpha\in Y$. Can anyone explain this to me?  Thanks.
80411
On $\mathbb{R}$ with the usual topology, generate an equivalence relation by $x\sim x$ for all $x$ in $\mathbb{R}$ and $x\sim 0$ for all $x$ in $[0,1]$. Prove $\mathbb{R}/\sim$ is homeomorphic to $\mathbb{R}$.
92345
Let $f: X \to Y$ be a bijective continuous function.  If $X$ is compact, and $Y$ is Hausdorff, then $f:X \to Y$ is a homeomorphism.   My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively.  I want to know if my counter examples are correct:  Let $f: X \to Y$ be the identity function.    (1)  Let $X=[0,1]$ with the standard topology and let $Y=[0,1]$ with the indiscrete topology.  Then $f: X \to Y$ is continuous and bijective but not a homeomorphism.  Since $U=(1/3,1/2)$ is open in $X$, but $(f^{-1})^{-1}(U)=f(U)$ is not open in $Y$.  (2) Let $X=Y=\mathbb{R}$.  Let $Y$ have the standard topology and $X$ have the discrete topology.  Let $U=\{x\}$ since open sets are singletons.  $f(^{-1})^{-1}(x)=f(x)$ is closed in $Y$ since $Y$ has the standard topology.  Hence $f$ is not a homeomorphism.  This is the old question My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively.  I want to know if my counter examples are correct:  Let $f$ be the identity function.  Then $f: X \to Y$ be a bijective continuous function.  If $X=[0,1]$ with the standard topology and $Y=[0,1]$ with the indiscrete topology we have that $f:X \to Y$ is not a homeomorphism.  Let $f: X \to Y$ be a bijective continuous function, $X=(0,1)$, and $Y=[0,1]$ both with the standard topology.  It's not a homeomorphism because $f^{-1}([0,1])$ is not connected, but $[0,1]$ is connected.  The first shows why Hausdoff is necessary for $Y$ and the second shows compactness is necessary for $X$.  Are both of the examples correct?  I have no confidence in my first example because I'm having problems showing $f^{-1}$ is not continuous.  However, I am confident in my second example.
46774
I've been studying introductory topology for a little bit now. I came across  video which explains open sets in a way I have never thought of. Even though the video is pretty elementary, I didn't know that open sets worked in this sense. If I were to 'imagine' moving toward the boundary of an open subset, I would just appear on the other side? Is this correct?  Is there any kind of intuitive way of thinking about the concept of an open set versus a closed set without just saying one is outlined with a dotted line versus a solid line. I ask this because I have never really understood why something like a domain is generally defined to be a set that is both open and connected. Why does it matter if it is open or closed? Is this just a formality? Or are open sets something that are special. I am curious because I would like to know if I have been thinking about open sets completely wrong.  Thanks.
57407
It is a Dugundji's exercise:  Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.  If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.  For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$.   Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.
122117
Baby Rudin gives the example of the set of all integers being not open if it is a subset of $\mathbb{R}^2$.  If we consider the set of integers in $\mathbb{R}$, is this set also not open? I can find a neighbourhood which will contain any point, $p$, however is it a requirement that a neighbourhood contains more than one point?  I'm trying to understand this fully and have searched through the various posts that have a slight relation and can not find out specifically how these take interior and isolated points into account and how these relate to openess.
115487
Is the 1 dimensional projective line homeomorphic to the circle?  If so, the circle is homeomorphic to itself with antipodal points identified  (very unintuitive).  I am missing something?
104747
Let $f:\mathbb R\to \mathbb R$ be a differentiable function such that $f(0)=  0$ and $f(1)= 1$ and $|f'(x)|&lt;2 ~ \forall x \in \mathbb R$, if $a$ and   $b$ are real numbers such that the set of possible values of $\displaystyle\int_0^1  f(x)dx $ is the open interval $(a,b)$, then $b-a$ is: ?   Attempt:   $$I = \int_0^1 1.f(x) dx$$  $$\implies I = 1 - \int_0^1 xf'(x)dx $$ (Integration by parts)  $$-2 &lt; f'(x) &lt; 2$$  $$\implies -2x &lt;xf'(x)&lt; 2x$$  for $x&gt;0$  $$\implies -1 &lt; \int_0^1 xf'(x) dx &lt; 1$$  Therefore $I_{max} = 2$ and $I_{min} = 0$  $\implies b- a = 2$ but answer given is $b-a = \dfrac 3 4$.  Please let me know my mistake, and the correct way to solve it.
29886
Question is :  Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $X\times X$. Then which of the following is true:   $X$ is not connected. $X$ is not compact $X$ is not homeomorphic to a subset of $\mathbb{R}$ None of the above.   I guess first two options are false.  We do have possibility that product of two connected spaces is connected.  So, $X\times X$ is connected if $X$ is connected. So I guess there is no problem.  We do have possibility that product of two compact spaces is compact.  So, $X\times X$ is compact if $X$ is compact. So I guess there is no problem.  I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.  So, only thing I have problem with is third option.  I could do nothing for that third option..  I would be thankful if some one can help me out to clear this.  Thank you :)
7440
Is there a symbol for "homeomorphic to"? I looked on Wikipedia, but it doesn't seem to mention one? Also, for isomorphism, is the symbol a squiggly line over an equals sign? What is the symbol with a squiggly line over just one horizontal line? Thanks.
92801
I want to ask about the homeomorphism between $\mathbb{Q}$, $\mathbb{Q}_{&gt;0}$: the rationals greater than $0$, and $\mathbb{Q}_{\geqslant 0}$: the rationals $\geqslant 0$?  For $\mathbb{Q}$ and $\mathbb{Q}_{&gt;0}$, I can use the back and forth map to have an order-isomorphism, and as a result, a homeomorphism. Is there a direct formula for the map?  It's very surprising that $\mathbb{Q}_{\geqslant 0}$ is homeomorphic to $\mathbb{Q}_{&gt;0}$ and $\mathbb{Q}$,  I don't know how to show they are homeomorphic.    Could anyone please help me with the homeomorphisms between them?
61068
Say a function's domain is $\{-\pi/2, \pi/2\}$. How should I interpret this interval?  It starts from where? To where? In what direction?  And if it starts from $3\pi/2$, would the next one be $-5\pi/3$.(because it starts from negative, $-\pi/2$).
56310
As the topic, how can I prove it?     Prove that (a,b) and [a,b) in R are not homeomorphic metric space.   I'm pretty weak on this section (homeomorphism) and I don't have any clue to figure it out.  Please help me, thank you.
123187
I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:     Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =  \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of   $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1&lt;y&lt;1\}$.      The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the   following reason: if $p$ is in the interval $I$, then there is no   adapted chart containing $p$, since any sufficiently small   neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely   many components.   Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.   I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.  I would appreciate any help!