How to prove that "If $X$ is a Hausdorff space equivalent to $\varDelta \left(x\right),$ it is a closed set in $X\times X$"？
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Let $X$ be a topological space. The diagonal of $X \times X$ is the subset $$D = \{(x,x)\in X\times X\mid x \in X\}.$$   Show that $X$ is Hausdorff if and only if $D$ is closed in $X \times X$.   First, I tried to show that $X \times X \setminus D$ is open using the fact that $X \times X$ is Hausdorff (because $X$ is Hausdorff), but I couldn't find an open set that contains a point outside $D$ and is disjoint to it...
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If $(X,d)$ is a complete metric space and $A$ is closed in $(X,d)$ then show that for each $x \in X$ there exists an element $a_0 \in A$ such that $d(x,A)=d(x,a_0).$   I tried this problem several times but always got stuck at some point or another. For instance, I initially thought about showing the existence of a Cauchy sequence $(a_n)$ in A such that $d(x,a_n) \to \operatorname{inf}\{d(x,a):a \in A\}=d(x,A)$ and then by completeness of $A$ and by continuity of the function $f(a)=d(x,a)$, to prove the existence of an element $a_0$ in $A$ such that $a_n \to a_0$ which would imply $d(x,a_n)=f(a_n) \to f(a_0)=d(x,a_0)=d(x,A)$. But the problem was that I could not construct such a Cauchy sequence in $A$. All my other attempts have similarly gone in vain.  If anybody has any answer, please reply. Thank you.
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A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.   Prove that every separable metric space has a countable base.   Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x&gt;0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)&lt;{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.   It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)&lt;\varepsilon_x/4$ and since $d(x, d_j)&lt;{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)&lt;{\varepsilon_x}/{4}+{\varepsilon_x}/{4}&lt;{\varepsilon_x}.$$  Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.  By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r&lt;{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.  Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls  arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.  Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.  Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.
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If $A$ is closed in $X$ and $B$ is closed in $Y$,   does it follow that $A \times B$ is closed in $X \times Y$ ?
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Let $X$ be a compact Hausdorff space. Show that the cone in $X$ is homeomorphic to the compactification of $X \times [0,1)$. If $A$ is closed in $X$, show that $X/A$ is homeomorphic to the compactification of $X\setminus A$.  I don't know even how to begin to solve this question, it's seems so hard, anyone could help me, please.
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If $A\times B$ is closed in $X\times Y$, then are $A$ and $B$ closed in $X$ and $Y$ respectively?
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I have troubles showing the following:  Let $(X,\rho)$ be a compact metric space and $F \subset X$ a closed subset. Prove that if diam $F &lt; \infty$, then there exist $x_{0}, y_{0} \in F$ such that diam $F= \rho(x_{0},y_{0})$.  It looks so trivial yet hard to find the trick.
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I'm thinking about Hausdorff spaces, and how mappings to Hausdorff spaces behave. Suppose I have an arbitrary (continuous) function $f:X \longrightarrow Y$, where $Y$ is a Hausdorff space (I think it is irrelevant for my question whether $X$ is Hausdorff or not, so I just consider it to be a topological space - if this is incorrect, please correct me!).  Can we say that the graph $$G_f=\{(x,f(x)) \in X \times Y\ : x \in X \}$$ is a closed subset of $X \times Y$? It seems quite obvious that it is the case, but I cannot see how to prove it. If anyone can offer a proof I'd be very interested. Regards.  EDIT 1  In response to Hennning Makholm:   I wasn't really aware of any variation in 'definition'; I guess I'm considering closed sets to be those with an open complement (though naturally this definition gives rise to other definitions, such as the subset equalling its closure etc.). For continuity of such a map, I would normally consider continuity to mean that $f^{-1}(V)$ is closed in $X$ whenever $V$ is closed in $Y$, though again definitions involving convergence of sequences and the notion that $f$ is continuous iff $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$ are also known to me.
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Suppose $X$ and $Y$ are topological vector spaces, $\dim (Y)&lt;\infty$, and $f:X\to Y$ is linear and surjective. Prove that $f$ is open and if the null space $N$ of $f$ is closed, then $f$ is also continuous.  The second assertion is easier since the quotient space $X/N$ has the universal mapping property.
107639
I know this question has been asked before, but I think I'm very close to a new solution and wanted to know if it is a viable approach.  Let $C(X)$ be the ring of continuous functions $X \rightarrow \mathbb{C}$ where $X$ is compact Hausdorff, and let $\mathfrak M$ be a maximal ideal.  We want to show that $\mathfrak M = I_{x_0} = \{ f \in C(X) : f(x_0) = 0\}$ for some $x_0$.  For each nonempty closed set $A$, we define the ideal $I_A = \{ f \in C(X) : f(x) = 0, \forall x \in A\}$.  Clearly $A \subset B$ implies $I_B \subset I_A$ (I also need to show that $I_B \subseteq I_A$ implies $A \subseteq B$, which I haven't done yet).  We let $\mathcal S$ be the set of ideals $I_A$ which are contained in $\mathfrak M$.  An ascending chain $I_{A_i} \in \mathcal S$ corresponds to a descending chain of closed sets $A_i$, whose intersection is nonempty because $X$ is compact.  Thus the union $J$ of the chain $I_{A_i}$ is an ideal which is contained in $I_{\bigcap A_i}$ (now we have to show that $I_{\bigcap A_i} \subseteq \mathfrak M$ to show that it is an upper bound for the chain, I haven't been able to do this).  So by Zorn's Lemma, $\mathcal S$ has a maximal element $I_D$ for some nonempty closed set $D$.  Now, I just want to show that $D$ is a singleton set $\{x_0\}$; then $I_D$ is a maximal ideal contained in $\mathfrak M$, whence $I_D = \mathfrak M$.  That will finish the proof.  Edit: Assuming I can fix the holes, I finished the problem (answered below).    Current problems with the proof:   (i) Need to show that $I_B \subseteq I_A$ implies $A \subseteq B$.  (ii) Need to finish the Zorn's lemma argument by showing that $I_{\bigcap\limits_i A_i} \subseteq \mathfrak M$.  Second edit: (i) follows from the Urysohn lemma.  A compact Hausdorff space is normal.  If $x \in A$, but not in $B$, then $\{x\}$ and $B$ are disjoint closed sets, so by Urysohn there exists a continuous function $f: X \rightarrow [0,1] \subseteq \mathbb{C}$ for which $f(x) = 1$ and $f(b) = 0$ for all $b \in B$.  Thus $f \in I_B$, but not in $I_A$.  So there is only one hole in the proof left.    Third edit: (ii) also follows from the Urysohn lemma. See the edited answer for details.
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I find it difficult to understand one of the properties and the definition of Hausdorff spaces.   The property states that, "If $X$ is Hausdorff then every finite set is closed. How is this possible keeping in mind that the definition states that for two distinct elements you can find an open neighborhood around them such that their intersection is empty. Thus, wouldn't the finite set be open because you can always find an open neighborhood around distinct elements of finite sets?   Can someone please explain?
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Let $A$ be closed in $X$ , where $X$ is a normal space, and $A$ is also a countable intersection of open sets. Prove that there exist a continuous function such that $$ \eqalign{   &amp; f\left( x \right) = 0\,\,\text{ if }\,x\, \in A  \cr    &amp; f\left( x \right) &gt; 0\,\,\text{ if }\,x \notin A \cr}  $$ EDITED: I mean intersection, sorry
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In the post  , it is argued that if $X$ is second countable and locally compact Hausdorff, then $C_{0}(X)$ is separable. Is it also true that $C(X)$ is separable under the same hypotheses?
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When a lattice $(L,\leqslant)$ is a lattice of open (or closed) sets of some topological space $(X,\tau)$? Which conditions have to be satisfied? We may assume that $X$ is $T_1$.
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I have almost solved the following problem but am stuck at the very end, can you help me finish it? Thank you for your help.       Let $n&lt;\omega$ and $t\in {}^n\omega$. We define $U_t=\{s\in {}^\omega\omega : t\subseteq s\}$. The family   $\mathcal B=\{U_t : t\in\bigcup {}^n\omega\}$ is a basis for a topology $\tau$ on ${}^\omega\omega$. This means that a set $X$   is in $\tau$ if and only if $X$ is a union of elements of $\mathcal B$.      EXERCISE 50 (PG): Show that $\langle {}^\omega\omega,\tau \rangle$ is a second countable completely metrizable space. Hint:    For $r,s\in {}^\omega\omega$ such that $r\ne s$, the $\varrho(r,s)=1/\min\{n : r(n)\ne s(n)\}$. Moreover, let $\varrho(r,r)=0$.    Show that $\varrho$ is a complete metric on ${}^\omega\omega$ that induces the topology $\tau$.     (i) Second countable: the set of $n$-tuples is countable since it is a finite union of countable sets.  (ii) complete with respect to $$ d(f,g) = \frac{1}{\min \{n \mid f(n) \neq g(n) \}}$$ Let $f_k$ be a Cauchy sequence. This means that the index at which $f_k, f_i$ differ gets larger as $i,k$ get larger. Let $f(n)$ denote the pointwise limit, that is, $\lim_{k \to \infty} f_k(n)$. Then $f(n) \in \omega$ for every $n$ since $f_k(n)$ is eventually constant. Hence $f \in \omega^\omega$.  (iii) $d$ induces the same topolgy as $U_t$:  $\tau_t \subset \tau_d$: Let $s \in U_t$. Then $B(s, \frac{1}{n+1})$ is an open ball contained in $U_t$ hence $U_t$ is open in $\tau_d$.  $\tau_t \supset \tau_d$: Let $B(s, \frac{1}{k})$ be an open balls. If $k &lt; n$, $s \in U_s \subset B(s, \frac{1}{k})$. But if $k \ge n$ then I'm stuck. The ball $B(s, \frac{1}{k})$ consists of all sequences that agree with $s$ on the first $k$ coordinates. So it is quite small. How do I make the $U_t$ small enough to fit in this ball? I have thought of intersection but since I'm only allowed to do finite intersections I don't see what to do.
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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?
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My question is  Let $A: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation, how can one prove that $\{Ay|y\geq 0\}$ is a closed set?  Here $y\geq 0$ means each component of $y$ $\geq 0$. I can simplify the question in two ways  Since the image of $A$ is a closed linear subspace of $\mathbb{R}^m$, by replacing $\mathbb{R}^m$ with this linear subspace, we may assume A is surjective.  After a change of basis of the codomain and a reordering of the basis of the domain, we may assume $A$ is of the form $[I|B]$   Any help or hints are appreciated, thank you.  Ideas: Let $v_i=Ae_i$, then my original statement is equivalent to saying the set $\{\sum a_iv_i |a_i\geq 0\}$ is a closed set. So I should now prove the following:  For any $v_1,...,v_k \in \mathbb{R}^n$, the set $S=\{\sum a_iv_i |a_i\geq 0\}$ is a closed subset of $\mathbb{R}^n$.   I suspect that the set $S$ is an intersection of half spaces, from this the closedness will be obvious. Equivalently, I suspect that for each $x\notin S$, there exists a half space $H$ (cut by a codimension 1 linear subspace)s.t. $S\subset H$. Is this statement true? I guess this should be true after drawing some examples.  New ideas: I am proving that the set $\{\sum a_iv_i|a_i\geq 0 \}$ is closed, and this is easy if all the $v_i$'s are indeed independent, so I would want to prove the following:  Let $S=\{$linearly independent subsets of $\{v_i\}\}$, then $\{\sum a_iv_i|a_i\geq 0\}=\cup_{T\subset S}\{$non negative linear combinations of $T\}$  And this union is closed since it is a finite union of closed set. Equivalently, let $x=\sum a_iv_i,a_i\geq 0$, I want to write $x=\sum b_ie_i$, where $b_i\geq 0$ and $\{e_i\}\subset \{v_i\}$ is linearly independent. My proof is as follows: I can replace $\{v_i\}$ with a smaller subset of itself s.t. $x$ cannot be expressed by non negative linear combinations of any strictly smaller subset of it. Now I claim that this set is linearly independent. Suppose not, then we can write $c_1v_1+c_2v_2+\cdots + c_nv_n=0$, with some $c_i$'s are positive(since we can multiply the whole expression by $-1$) and I assume that $c_i&gt;0$ for $i=1,2,\dots ,k$ and $c_i \leq 0$ for $i&gt;k$ Now I assume $b=a_1/c_1=\min _{i=1,2,\dots ,k}{a_i/c_i}$, Then I have $$x=\sum a_iv_i-b(\sum c_iv_i)=                      (a_2-bc_2)v_2     +(a_3-bc_3)+\cdots + (a_n-bc_n)v_n$$ Which is a non negative linear combinations of fewer vectors, contradicts to my assumption and hence the minimal set $\{v_i\}$ is linearly independent.