In a lecture on Baire's Theorem (for complete metric spaces), I gave, for a rather advanced undergraduate class in Real Analysis (covering the theory of metric spaces and elements of general topology), I realised that all the applications I provided were either straight-forward or could be proved using more elementary tools. For example:  a. $\mathbb Q$ is not complete.  b. $\mathbb R$ is uncountable. (Otherwise, if $\mathbb R=\{x_n\}_{n\in\mathbb N}$, then $\bigcap_{n\in\mathbb N} (\mathbb R\setminus\{x_n\})$ would be dense and empty.)   c. A complete metric, with the property that every point is an accumulation point, is uncountable.  Could you suggest more interesting applications? They do not have to be too easy to prove.
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I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power.  Here's the theorem (with proof) and two applications:    (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.  Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 &gt; 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$.   Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 &gt; 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$    Here is one application (taken from ):  Claim: $[0,1]$ contains uncountably many elements.  Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in (0,1)} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable.     And here is another one (taken from ):  Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.  Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem."
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Show that $\mathbb R$ is cardinally larger than any separable metric space S.  I have been trying to solve this on my own. My idea was to start by mapping the open balls of positive rational radius around the points in the dense set of S to the corresponding in $\mathbb R$. Now any ${x}\subset S$ can be written as a limit of an appropriate sequence of open balls, which would correspond to a similar limit in $\mathbb R$, in case the limit exists.  Then I would need to show that different limits are obtained in $\mathbb R$ for different $x\in S$. This is a step I am not sure how to show.   Alternatively, one could show that all possible sequences of rationals have no greater cardinality than reals. What would be a good way to show this? I know that all reals can be written as some convergent sequence of rationals, but I am looking for something slightly different here.  I did not find this useful as it does not fill in the details:
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This question is related to Exercise 2.15 in Baby Rudin, in which the reader is instructed to "[s]how that Theorem 2.36 and its Corollary become false (in $R^1$, for example) if the word 'compact' is replaced by 'closed' or by 'bounded.'"  The relevant theorem and corollary are as follows:  Theorem 2.36: If $\{K_{α}\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{α}\}$ is nonempty, then   $\{K_{α}\}$ is nonempty.  Corollary: If $\{K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n}$ contains $K_{n+1},  (n = 1, 2, 3, ...)$, then   $K_n$ is not empty.  My Question: I realize this question and similar ones have been asked elsewhere on MSE, but I've yet to see a straightforward (to me, at least) answer to my question specifically. Usually, for the "closed" part of the exercise, the following collection of closed sets, $\{K_n \}_{n \in \mathbb{N}}$ is used:  $$K_n = [n, \infty), n \in \mathbb{N}.$$  For, the intersection of every finite subcollection of ${K_n}$ is clearly nonempty. Indeed,  $$ \bigcap_{i=1}^{m} K_{n_{i}} = [N, \infty) $$  where $N = \max\{n_1, n_2, ..., n_m\}$.  Then, the assertion is made that   $$ \bigcap_{n=1}^{\infty} K_n = \emptyset. $$  However, I don't see how the same argument can't simply be used, i.e., that, for example, $N \in  \bigcap_{n=1}^{\infty} K_n$ (or maybe, say, $n+1$). Intuitively, this certainly makes sense, I'd say. For, given the archimedean property, it seems that as $n$ approaches infinity, we can always find an $l$ in each increasingly smaller set such that $n &lt; l &lt; \infty$; and so $l \in  \bigcap_{n=1}^{\infty} K_n$.  What is it exactly that I'm misunderstanding or overlooking here?
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Can someone please check whether the following simple proof is "mathematical"? Is it correct, complete, rigid? Can it be simplified? I'm a complete autodidact so I'm looking for someone to give me feedback to gain experience in writing proofs... This is also my first question on MSE.  The proposition:  Let $(X, d)$ be a metric space and $x_n \to x$ where each $x_n \in X$ and $x \in X$. Let $A$ be the subset of $X$ which consists of $x$ and all of the points $x_n$. Prove that $A$ is closed in $(X, d)$.  My tentative to prove this:  We first show that all infinite sequences in $A$ converge to $x$: Let $y \in X$, $y \ne x$. Then there is some open ball $B_\epsilon(x)$ with $\epsilon &lt; d(x,y)$ containing all but finitely many elements of $A$. As $y \notin B_\epsilon(x)$ there can be no infinite sequence in $A$ converging to $y$. Consequently all infinite series in $A$ converge to a point in $A$ which therefore must be a closed set.  Edited: As rightly pointed out in the comments, I should have written in the first sentence "...sequences with infinitely many distinct terms and which converge to some point of $X$" and the last sentence should be "Consequently all infinite sequences...".
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I believe that the set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$, the . However, I read in the book &quot;Metric spaces&quot; by Ó Searcóid that set of all $[0, 1]\to\mathbb{R}$ continuous functions is greater than $\mathfrak c$:  &quot;It is demonstrated in many textbooks that $\mathbb{Q}$ is countable, that $\mathbb{R}$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions deﬁned on  $[0,1]$ is of a greater cardinality than $\mathbb{R}$, and that there are sets of greater and greater cardinality.&quot;  I understand that (via composition with the continuous function $\tan$ or $\arctan$) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this?
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Suppose $(S,d)$ is a non-compact metric space. Is it possible to construct an unbounded continuous function from $S$ to $\mathbb{R}$? If it is possible, please show the construction method :)  Here's my attempt to solve this problem: Because $S$ is not compact, we can find a neighborhood assignment function $N(x)$ so that a finite cover is not possible. Then each time we can find an uncovered point from the union of neighborhoods before, and change the function by $f(x):=max\{f(x),\text{a function which equals to the indice of this point at that point and gradually vanishes at further points from the selected point}\}$. However I'm concerned that while this function is unbounded, there may be some invalid points for this function (goes to infinity)...  EDIT: As I found out there is a duplicate question, however the elementary solution is not presented in that question, so I still wish to know about this problem. :)
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Let $(X,d)$ be a metric space and let $S$ be the set of Cauchy sequences in $S$. Define "~" in $S$ by "$\{s_k\}$ ~ $\{t_k\}$" to mean that $d(s_k,t_k) \to 0$ as $k \to \infty$. Let $X'$ denote the set of equivalence classes of $S$ and let $s'$ denote the equivalence class of $s = \{s_k\}$.   I know $\phi(s',t') = \lim\limits_{k \to \infty} d(s_k,t_k)$ is a metric on $X'$.   Show $(X',\phi)$ is complete.  For $x \in X$, define $x'$ to be the equivalence class of the constant sequence ${x,x,...}$. Show that the function $x \to x'$ is an isometry of $X$ onto a dense subset of $X$.   If anyone can share a proof, or steer me in the right direction, maybe tell me what it is that I need to prove specifically, that would be great.
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Let $\overline{B_d (x, r)}$ be the closure of an open ball, and let $\hat B_d (x, r)$ be a closed ball. Show that in any metric space $(X, d)$, $\overline{B_d (x, r)} \subseteq \hat B_d (x, r)$ for all $x \in X$ and $r &gt; 0$.    I used the definition of closure: $\overline{B}$ = $B \bigcup B'$. If $x \in \overline{B}$, then $x \in B$ or $x \in \overline{B}$, or both. If $x \in B$, then clearly $x \in \hat B$ since $B \subseteq \hat B$. But, if $x \in B'$, then $x$ is an accumulation point. Since $\hat B$ is closed, it contains all accumulation points of $B$. Thus, $x \in \hat B$.  Is this approach correct, or at least in the right direction? Should I be using the closure definition, or try another one? Thank you.
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I know this proof is somewhat similar, or related to the Baire's Category Theorem but I can't seem to figure out how to do it.  The Baire Category theorem asserts that if X is a complete metric space or a locally compact Hausdorff space (my case), then the complement of a countable union of nowhere dense sets is always nonempty.  How can I use this on the proof?
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I wish to show that any metrizable space $(X,\mathcal{T})$ is Hausdorff  Proof attempt:  Let $d$ be the metric that generates the topology on $X$. Pick two points $x,y \in X$, we wish to produce two disjoint open sets that separates $x,y$.  Let $B_\epsilon(x)$ and $B_\delta(y)$ be two metric balls containing $x,y$ respectively. Suppose that  $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$       Stuck Here: Hmm...How should I adjust $\epsilon, \delta$ so that these   balls are separated?   Idea: Reduce $\epsilon$ by half. If they are still intersecting...reduce $\delta$ by half. Continue ad infinitum         Is there a more satisfying solution i.e. closed form expression for   reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!
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My question: let $f : X \to \Bbb R$ be a non-constant continuous function on a connected metric space and assume that $f(X)$ is uncountable; then $X$ is uncountable.  We know continuous image of a connected metric space is connected. Since connected space in $\Bbb R$ are intervals and the function is non-constant continuous function so $f(X)$ is an interval in $\Bbb R$ and hence uncountable.  Is the proof correct?? How does it follows that $X$ is uncountable?
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This is not a very well-defined question.  Are there any standard constructions of metric spaces, parameterized by real-valued $n \ge 1$, such that:   When $n$ is an integer, the metric space is precisely $\mathbb{R}^n$. When $n$ is non-integer, the metric space can be seen as a reasonable generalization of $\mathbb{R}^n$.  For example, perhaps it has  of $n$.   Alternately, a non-existence result that you can't maintain some of the important properties of $\mathbb{R}^n$ in a generalization like this would be interesting to me.
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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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I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d)$. Let $d$ be complete. Does there exist an equivalent metric $d'$ in $X$ which is incomplete? When $d$ is compact, such $d'$ does not exist (Since compactness implies completeness). So answer me when $d$ is noncompact.
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I know that every separable metric space has a countable base.  I was wondering if we can get a countable dense subset from every metric space that has a countable base.  Thank you very much!!
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Is  the  space $\mathbb{N}$  is a  baire  spaces ?  My attempts :   No ,  It  will not form a  baires  spaces ,  let $C_n= \mathbb{N}\setminus\{n\}$. This is open (since $\{n\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{N}$) and dense, since every open ball with center in $\mathbb{n}$ intersects ${C_n}$. But  $$\bigcap_{n\in\mathbb{N}} {C}_n = \emptyset$$  So $\mathbb{N}$ is not a baires  spaces because for  Baire Space the intersection of a countable family of open dense sets must be dense.   Is  Its correct or not ?
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I've seen a few questions already posted on here, but they all deal with perfect subsets and being Hausdorff, both of which are topics we haven't covered yet.  I know that a point a is isolated if $\{a\}$ is open (aka there exists $\epsilon &gt; 0$ s.t. $B_{\epsilon}(a) = \{a\}$.  I am supposed to consider singletons and then use Baire Category Theorem.  Could someone help me with where to start?