00:00 Introduction

00:40 Compactness in R^n

01:46 Definition: sequentially compact

04:00 Examples

05:52 Proposition: compact implies closed and bounded

07:14 Proof

1 00:00:00,360 –> 00:00:02,059 Hello and welcome back to

2 00:00:02,069 –> 00:00:03,279 functional analysis.

3 00:00:03,440 –> 00:00:04,659 And as always, I want to

4 00:00:04,670 –> 00:00:06,059 thank all the nice people

5 00:00:06,070 –> 00:00:07,619 that make this video possible

6 00:00:07,630 –> 00:00:09,100 by supporting me on steady

7 00:00:09,109 –> 00:00:09,899 or paypal.

8 00:00:10,430 –> 00:00:11,659 Now, we have reached part

9 00:00:11,670 –> 00:00:13,340 16 in our course.

10 00:00:13,350 –> 00:00:15,029 And we’ll talk about compactness

11 00:00:15,039 –> 00:00:15,520 today.

12 00:00:16,430 –> 00:00:17,969 The word compact you find

13 00:00:17,979 –> 00:00:19,270 very often in functional

14 00:00:19,280 –> 00:00:20,000 analysis.

15 00:00:20,010 –> 00:00:21,510 However, here we first go

16 00:00:21,520 –> 00:00:23,000 back to metric spaces.

17 00:00:23,870 –> 00:00:25,409 It’s a property we define

18 00:00:25,420 –> 00:00:27,079 for subsets in a matrix

19 00:00:27,090 –> 00:00:28,370 space and it should

20 00:00:28,379 –> 00:00:29,899 extend the finite.

21 00:00:29,959 –> 00:00:31,629 So a finite

22 00:00:31,639 –> 00:00:32,810 set should always be

23 00:00:32,819 –> 00:00:33,650 compact.

24 00:00:34,020 –> 00:00:35,259 Then for an infinite set,

25 00:00:35,270 –> 00:00:36,689 the compactness would say

26 00:00:36,700 –> 00:00:38,389 that your movement is still

27 00:00:38,400 –> 00:00:40,060 restricted in a similar way

28 00:00:40,610 –> 00:00:41,880 to get an idea of what this

29 00:00:41,889 –> 00:00:42,930 actually means.

30 00:00:42,939 –> 00:00:44,779 Let’s look at a subset in

31 00:00:44,790 –> 00:00:45,389 RN.

32 00:00:46,299 –> 00:00:47,529 Of course, we have our N

33 00:00:47,540 –> 00:00:49,080 with our usual Euclidean

34 00:00:49,090 –> 00:00:49,259 me.

35 00:00:50,229 –> 00:00:51,409 Now, you know what it means

36 00:00:51,419 –> 00:00:53,159 to say that A is a closed

37 00:00:53,169 –> 00:00:54,900 set, the

38 00:00:54,909 –> 00:00:56,380 whole boundary should belong

39 00:00:56,389 –> 00:00:57,750 to the whole set itself,

40 00:00:57,759 –> 00:00:59,470 which means you can’t leave

41 00:00:59,479 –> 00:01:01,029 it from the inside by using

42 00:01:01,040 –> 00:01:01,709 a sequence.

43 00:01:02,520 –> 00:01:04,470 So you see this is one restriction

44 00:01:04,480 –> 00:01:05,870 you have for your movement.

45 00:01:06,489 –> 00:01:07,739 And another one comes in.

46 00:01:07,750 –> 00:01:09,250 If you look at a bounded

47 00:01:09,260 –> 00:01:11,010 set, this

48 00:01:11,019 –> 00:01:12,239 simply means that there is

49 00:01:12,250 –> 00:01:13,699 a maximum distance you can

50 00:01:13,709 –> 00:01:15,500 go, you are not allowed to

51 00:01:15,510 –> 00:01:16,599 go to infinity.

52 00:01:17,349 –> 00:01:18,879 Now, you might already know

53 00:01:18,940 –> 00:01:20,489 these two things together

54 00:01:20,540 –> 00:01:21,930 is what we can shorten to

55 00:01:21,940 –> 00:01:22,849 compact.

56 00:01:23,400 –> 00:01:25,150 However, please be careful.

57 00:01:25,160 –> 00:01:26,569 Actually, this is not a

58 00:01:26,580 –> 00:01:27,430 definition.

59 00:01:27,480 –> 00:01:29,330 This is a fact we only have

60 00:01:29,339 –> 00:01:31,239 in RN or if you

61 00:01:31,250 –> 00:01:33,019 want also in CN

62 00:01:33,029 –> 00:01:34,209 together with this standard

63 00:01:34,220 –> 00:01:34,709 metric.

64 00:01:35,660 –> 00:01:36,849 Therefore, the correct question

65 00:01:36,860 –> 00:01:38,199 here would be how can we

66 00:01:38,209 –> 00:01:39,639 define compact in

67 00:01:39,650 –> 00:01:41,199 general such that we get

68 00:01:41,209 –> 00:01:42,800 the correct notion for infinite

69 00:01:42,809 –> 00:01:43,230 sets?

70 00:01:44,000 –> 00:01:45,800 Of course, this is now what

71 00:01:45,809 –> 00:01:46,480 we will do.

72 00:01:47,680 –> 00:01:49,389 So let’s fix a metric space

73 00:01:49,400 –> 00:01:51,300 XD in the

74 00:01:51,309 –> 00:01:52,830 same way as before, let’s

75 00:01:52,839 –> 00:01:54,319 consider a subset of this

76 00:01:54,330 –> 00:01:54,900 space.

77 00:01:55,809 –> 00:01:57,650 And now we define this as

78 00:01:57,660 –> 00:01:59,330 a compact set or more

79 00:01:59,339 –> 00:02:01,029 concretely as a sequentially

80 00:02:01,040 –> 00:02:01,910 compact set.

81 00:02:02,849 –> 00:02:04,510 Maybe you already recognize

82 00:02:04,519 –> 00:02:05,900 the same thing that happened

83 00:02:05,910 –> 00:02:07,489 in the continuity definition.

84 00:02:08,187 –> 00:02:09,598 There are two terms you could

85 00:02:09,608 –> 00:02:11,598 define one with sequences

86 00:02:11,608 –> 00:02:12,938 and the other one without

87 00:02:12,949 –> 00:02:13,768 sequences.

88 00:02:14,218 –> 00:02:15,848 And then the result for metric

89 00:02:15,858 –> 00:02:17,529 spaces is they are

90 00:02:17,539 –> 00:02:18,158 equivalent.

91 00:02:18,809 –> 00:02:19,990 It’s the same thing here,

92 00:02:20,000 –> 00:02:21,289 but I don’t want to bother

93 00:02:21,300 –> 00:02:22,850 you with the actual compact

94 00:02:22,860 –> 00:02:23,389 definition.

95 00:02:24,059 –> 00:02:25,580 The term sequentially compact

96 00:02:25,589 –> 00:02:27,399 is just easier to understand

97 00:02:27,520 –> 00:02:29,130 and we just call it compact.

98 00:02:29,139 –> 00:02:31,000 Now, however, please

99 00:02:31,009 –> 00:02:32,240 keep in mind when you’re

100 00:02:32,250 –> 00:02:33,940 not working in metric spaces.

101 00:02:33,949 –> 00:02:35,520 But in general topological

102 00:02:35,529 –> 00:02:37,330 spaces, you have two terms

103 00:02:37,339 –> 00:02:38,639 that are not equivalent

104 00:02:39,229 –> 00:02:39,679 for us.

105 00:02:39,690 –> 00:02:40,720 It’s not a problem here.

106 00:02:40,729 –> 00:02:42,720 We call a compact if

107 00:02:42,729 –> 00:02:44,130 for each sequence in a,

108 00:02:44,860 –> 00:02:46,479 we find a convergent

109 00:02:46,490 –> 00:02:48,410 subsequence where

110 00:02:48,419 –> 00:02:49,720 the limit X tilde

111 00:02:49,729 –> 00:02:51,410 still lies in the set

112 00:02:51,419 –> 00:02:51,759 A.

113 00:02:52,479 –> 00:02:53,690 So you see there’s a lot

114 00:02:53,699 –> 00:02:55,440 going on here, you start

115 00:02:55,449 –> 00:02:57,250 with a sequence, then you

116 00:02:57,259 –> 00:02:58,490 omit some members,

117 00:02:59,059 –> 00:03:00,070 which means that you get

118 00:03:00,080 –> 00:03:01,839 a subsequence, which is

119 00:03:01,850 –> 00:03:03,600 also convergent where the

120 00:03:03,610 –> 00:03:05,440 limit still lies in a,

121 00:03:05,970 –> 00:03:06,710 in the picture.

122 00:03:06,720 –> 00:03:08,199 This would mean inside a

123 00:03:08,210 –> 00:03:09,529 set A, we have

124 00:03:09,539 –> 00:03:11,229 infinitely many points

125 00:03:11,240 –> 00:03:12,630 given by the sequence

126 00:03:12,639 –> 00:03:13,270 XN.

127 00:03:14,220 –> 00:03:15,449 Of course, a lot can happen

128 00:03:15,460 –> 00:03:16,490 for this sequence.

129 00:03:16,500 –> 00:03:18,330 But compact now means

130 00:03:18,339 –> 00:03:20,229 that you can choose infinitely

131 00:03:20,240 –> 00:03:21,970 many points in the sequence

132 00:03:21,979 –> 00:03:23,250 such that you get a limit

133 00:03:23,259 –> 00:03:23,690 out.

134 00:03:24,279 –> 00:03:25,860 For example, here, only one

135 00:03:25,869 –> 00:03:27,559 part remains where we have

136 00:03:27,570 –> 00:03:29,220 convergence to a point X

137 00:03:29,240 –> 00:03:29,779 tilde.

138 00:03:30,250 –> 00:03:31,850 And this should work no matter

139 00:03:31,860 –> 00:03:33,160 which sequence you started

140 00:03:33,169 –> 00:03:33,490 with.

141 00:03:34,240 –> 00:03:35,699 Now, if you go back to our

142 00:03:35,729 –> 00:03:37,039 end with this knowledge,

143 00:03:37,179 –> 00:03:38,460 you immediately see this

144 00:03:38,470 –> 00:03:40,089 excludes unbounded

145 00:03:40,660 –> 00:03:42,020 because if you tend to

146 00:03:42,029 –> 00:03:43,899 infinity, you can’t find

147 00:03:43,910 –> 00:03:45,690 such a convergence subsequence.

148 00:03:46,259 –> 00:03:48,169 Also the set has to be closed.

149 00:03:48,179 –> 00:03:49,770 Otherwise you can just choose

150 00:03:49,779 –> 00:03:51,539 a sequence that has a limit

151 00:03:51,550 –> 00:03:52,699 outside of the set.

152 00:03:52,710 –> 00:03:53,009 A.

153 00:03:53,149 –> 00:03:54,759 Then of course, you can’t

154 00:03:54,770 –> 00:03:56,039 add a new limit with a

155 00:03:56,050 –> 00:03:56,809 subsequence.

156 00:03:57,789 –> 00:03:58,270 OK.

157 00:03:58,279 –> 00:03:59,690 Next, let’s look at some

158 00:03:59,699 –> 00:04:00,410 examples.

159 00:04:01,369 –> 00:04:02,669 For the first example, let’s

160 00:04:02,679 –> 00:04:04,300 choose a more concrete one

161 00:04:04,309 –> 00:04:05,600 than the one from above.

162 00:04:06,910 –> 00:04:08,039 So this should be just the

163 00:04:08,050 –> 00:04:09,699 real number line together

164 00:04:09,710 –> 00:04:11,059 with the standard metric,

165 00:04:11,070 –> 00:04:12,580 the Euclidean metric which

166 00:04:12,589 –> 00:04:13,800 is given by the absolute

167 00:04:13,809 –> 00:04:14,259 value.

168 00:04:14,910 –> 00:04:16,730 And the set A should be given

169 00:04:16,738 –> 00:04:18,329 as the unit interval.

170 00:04:18,910 –> 00:04:20,428 Now, in this one dimensional

171 00:04:20,440 –> 00:04:21,928 case, you should see that

172 00:04:21,940 –> 00:04:22,970 the formulation with the

173 00:04:22,980 –> 00:04:24,649 sequences and the convergent

174 00:04:24,660 –> 00:04:26,040 subsequence is

175 00:04:26,049 –> 00:04:27,730 exactly what you learn in

176 00:04:27,739 –> 00:04:28,529 analysis.

177 00:04:28,989 –> 00:04:30,519 It’s the famous Bolzano

178 00:04:31,109 –> 00:04:31,769 Vastra theorem.

179 00:04:32,769 –> 00:04:34,200 Then with this in a tool

180 00:04:34,209 –> 00:04:35,970 set, one can easily show

181 00:04:35,980 –> 00:04:37,730 the equality here for RN

182 00:04:37,739 –> 00:04:39,359 or CN with the standard

183 00:04:39,369 –> 00:04:39,859 metric.

184 00:04:40,670 –> 00:04:42,160 Therefore, let’s now look

185 00:04:42,170 –> 00:04:43,529 at the same set together

186 00:04:43,540 –> 00:04:44,609 with another metric.

187 00:04:45,489 –> 00:04:46,709 So again, the real number

188 00:04:46,720 –> 00:04:48,679 line, but now with the discrete

189 00:04:48,690 –> 00:04:50,640 metric, now you might

190 00:04:50,649 –> 00:04:52,070 already guess in this

191 00:04:52,079 –> 00:04:53,850 case, the unit interval is

192 00:04:53,859 –> 00:04:55,049 not compact.

193 00:04:55,510 –> 00:04:56,880 In order to show this, you

194 00:04:56,890 –> 00:04:58,850 just have to find one sequence

195 00:04:58,859 –> 00:05:00,170 which has no convergence

196 00:05:00,179 –> 00:05:01,440 subsequence at all.

197 00:05:02,200 –> 00:05:04,019 Hence, let’s simply do that.

198 00:05:04,869 –> 00:05:05,279 OK.

199 00:05:05,290 –> 00:05:06,739 So let’s choose to sequence

200 00:05:06,750 –> 00:05:07,989 XN by setting

201 00:05:08,000 –> 00:05:09,760 XN as one over

202 00:05:09,769 –> 00:05:11,410 N, it’s a

203 00:05:11,420 –> 00:05:12,959 sequence inside the unit

204 00:05:12,970 –> 00:05:14,350 interval but it’s not

205 00:05:14,359 –> 00:05:15,029 constant.

206 00:05:15,940 –> 00:05:17,369 Hence, we already know what

207 00:05:17,380 –> 00:05:18,920 the distance is between two

208 00:05:18,929 –> 00:05:20,329 members of the sequence

209 00:05:20,720 –> 00:05:22,220 we are in the discrete metric

210 00:05:22,230 –> 00:05:22,799 space.

211 00:05:22,809 –> 00:05:24,670 So we get always one or

212 00:05:24,679 –> 00:05:25,929 two different points.

213 00:05:26,470 –> 00:05:28,239 However, this now means that

214 00:05:28,250 –> 00:05:30,149 we could omit a lot of members

215 00:05:30,160 –> 00:05:31,899 in the sequence, but we never

216 00:05:31,910 –> 00:05:33,579 get the one down to zero

217 00:05:33,589 –> 00:05:33,920 here.

218 00:05:34,750 –> 00:05:35,959 In other words, there can’t

219 00:05:35,970 –> 00:05:37,959 be any convergent subsequence.

220 00:05:38,799 –> 00:05:40,609 In conclusion, the unit interval

221 00:05:40,619 –> 00:05:42,239 is not compact with

222 00:05:42,250 –> 00:05:43,839 respect to the discrete metric.

223 00:05:44,690 –> 00:05:44,970 OK.

224 00:05:44,980 –> 00:05:46,570 So please keep that in mind,

225 00:05:46,809 –> 00:05:48,220 always emphasize the

226 00:05:48,230 –> 00:05:49,690 corresponding metric when

227 00:05:49,700 –> 00:05:51,420 talking about compact sets.

228 00:05:52,200 –> 00:05:53,600 Now for the end of the video,

229 00:05:53,609 –> 00:05:54,880 I want to show you a nice

230 00:05:54,890 –> 00:05:56,809 proposition which generalizes

231 00:05:56,820 –> 00:05:58,079 the results from above,

232 00:05:58,779 –> 00:06:00,029 we have our metric space

233 00:06:00,040 –> 00:06:01,869 XD and a compact

234 00:06:01,880 –> 00:06:03,730 set A then

235 00:06:03,739 –> 00:06:04,790 we can conclude.

236 00:06:04,799 –> 00:06:06,470 So it’s an implication not

237 00:06:06,480 –> 00:06:07,309 an equivalence

238 00:06:08,290 –> 00:06:10,179 that A is closed

239 00:06:10,190 –> 00:06:11,309 and bounded.

240 00:06:11,660 –> 00:06:13,390 So in general, we still have

241 00:06:13,399 –> 00:06:15,049 the result that closeness

242 00:06:15,059 –> 00:06:16,910 and bounded is needed

243 00:06:16,920 –> 00:06:18,179 for the compactness.

244 00:06:18,720 –> 00:06:20,109 It’s not the other way around

245 00:06:20,119 –> 00:06:21,790 compactness in general is

246 00:06:21,799 –> 00:06:22,440 more.

247 00:06:22,450 –> 00:06:24,070 But this can still be helpful

248 00:06:24,850 –> 00:06:26,359 before writing down the proof,

249 00:06:26,369 –> 00:06:27,799 we should first define what

250 00:06:27,809 –> 00:06:29,399 we mean by a bounded set

251 00:06:29,410 –> 00:06:30,839 inside the metric space.

252 00:06:31,700 –> 00:06:32,859 Of course, this is not so

253 00:06:32,869 –> 00:06:34,380 surprising if you have your

254 00:06:34,390 –> 00:06:36,190 set a here, it

255 00:06:36,200 –> 00:06:37,410 simply means that you can’t

256 00:06:37,420 –> 00:06:39,290 find a large enough epsilon

257 00:06:39,299 –> 00:06:41,029 ball such that the whole

258 00:06:41,040 –> 00:06:42,640 set lies inside this

259 00:06:42,649 –> 00:06:43,170 ball.

260 00:06:43,679 –> 00:06:44,989 So we just have to find a

261 00:06:45,000 –> 00:06:46,489 middle point X

262 00:06:46,649 –> 00:06:48,279 and a radius epsilon.

263 00:06:48,899 –> 00:06:50,170 And then we want that the

264 00:06:50,179 –> 00:06:52,079 open Epsom ball around

265 00:06:52,089 –> 00:06:53,959 X is a superset for A

266 00:06:54,839 –> 00:06:55,989 and this is what we mean

267 00:06:56,000 –> 00:06:57,339 when we say bounded,

268 00:06:58,010 –> 00:06:59,230 it simply means that there

269 00:06:59,239 –> 00:07:01,049 is a constant such that any

270 00:07:01,059 –> 00:07:02,790 two points have distance

271 00:07:02,799 –> 00:07:04,329 less than this constant.

272 00:07:04,980 –> 00:07:06,679 And here you see this constant

273 00:07:06,690 –> 00:07:08,350 would be just too epsilon.

274 00:07:09,149 –> 00:07:09,519 OK.

275 00:07:09,529 –> 00:07:10,989 By knowing all the terms

276 00:07:11,000 –> 00:07:12,359 in the proposition, we now

277 00:07:12,369 –> 00:07:13,760 can write down the proof

278 00:07:14,049 –> 00:07:15,779 starting from a compact set,

279 00:07:15,790 –> 00:07:17,230 we have to do two things.

280 00:07:17,380 –> 00:07:18,980 First, we want to show that

281 00:07:18,989 –> 00:07:19,959 A is closed.

282 00:07:19,970 –> 00:07:21,500 So let’s choose a convergence

283 00:07:21,510 –> 00:07:22,429 sequence in a.

284 00:07:22,910 –> 00:07:24,019 Hence we know there is a

285 00:07:24,029 –> 00:07:25,480 limit X tilde in

286 00:07:25,489 –> 00:07:26,049 X.

287 00:07:26,519 –> 00:07:28,290 Now we call, if we can show

288 00:07:28,299 –> 00:07:29,760 that X tilde is actually

289 00:07:29,769 –> 00:07:31,519 in a, then A is closed

290 00:07:32,209 –> 00:07:33,750 using the compactness, we

291 00:07:33,760 –> 00:07:34,890 immediately get the result

292 00:07:34,899 –> 00:07:35,989 for a subsequence.

293 00:07:36,470 –> 00:07:37,750 Hence, we have a convergent

294 00:07:37,760 –> 00:07:39,660 subsequence with limit

295 00:07:39,670 –> 00:07:40,230 in A.

296 00:07:40,239 –> 00:07:41,630 And maybe we call this limit

297 00:07:41,640 –> 00:07:42,779 X tilde tilde.

298 00:07:43,779 –> 00:07:45,549 Of course, this is not necessary

299 00:07:45,559 –> 00:07:47,269 because we know we already

300 00:07:47,279 –> 00:07:48,670 started with a convergence

301 00:07:48,679 –> 00:07:49,250 sequence.

302 00:07:49,779 –> 00:07:51,329 And we know the limit is

303 00:07:51,339 –> 00:07:52,250 always unique.

304 00:07:52,260 –> 00:07:53,730 You can’t find another limit

305 00:07:53,739 –> 00:07:54,739 with a subsequence.

306 00:07:55,380 –> 00:07:57,130 Hence X tilde and X tilde

307 00:07:57,290 –> 00:07:59,049 tilde do coincide.

308 00:07:59,059 –> 00:08:00,690 So we know X tilde lies

309 00:08:00,700 –> 00:08:02,589 also in a, this is

310 00:08:02,600 –> 00:08:03,790 exactly what we wanted to

311 00:08:03,799 –> 00:08:04,290 show.

312 00:08:04,399 –> 00:08:06,109 Therefore A is closed.

313 00:08:06,809 –> 00:08:07,239 OK.

314 00:08:07,250 –> 00:08:08,779 Then the next step is showing

315 00:08:08,790 –> 00:08:10,019 that A is bounded.

316 00:08:10,480 –> 00:08:12,000 We do this with a proof by

317 00:08:12,010 –> 00:08:12,839 con interposition.

318 00:08:13,589 –> 00:08:14,920 This means that we start

319 00:08:14,929 –> 00:08:16,149 with a being not

320 00:08:16,160 –> 00:08:18,140 bounded and we want to end

321 00:08:18,149 –> 00:08:19,500 with a being not

322 00:08:19,510 –> 00:08:20,230 compact.

323 00:08:20,890 –> 00:08:22,869 Now, if A is not bounded,

324 00:08:22,880 –> 00:08:24,790 we know that we can find

325 00:08:24,799 –> 00:08:26,720 distances as large as

326 00:08:26,730 –> 00:08:27,230 we want.

327 00:08:27,950 –> 00:08:29,890 Hence, for given point A

328 00:08:29,899 –> 00:08:31,609 in the set A, we

329 00:08:31,619 –> 00:08:33,419 find infinitely many points

330 00:08:33,429 –> 00:08:34,969 XN in A

331 00:08:34,979 –> 00:08:36,690 such that the distance to

332 00:08:36,700 –> 00:08:38,340 A is larger than the

333 00:08:38,349 –> 00:08:39,450 natural number N.

334 00:08:40,890 –> 00:08:42,719 With this, we have one sequence

335 00:08:42,729 –> 00:08:44,409 XN in A where we can look

336 00:08:44,419 –> 00:08:45,409 at all possible

337 00:08:45,419 –> 00:08:46,419 subsequence.

338 00:08:46,840 –> 00:08:48,770 Now, if A was compact,

339 00:08:48,780 –> 00:08:50,619 we would find a subsequence

340 00:08:50,630 –> 00:08:51,419 with a limit.

341 00:08:52,260 –> 00:08:53,520 Therefore, let’s look at

342 00:08:53,530 –> 00:08:54,929 all possible limit points

343 00:08:54,940 –> 00:08:55,549 as well.

344 00:08:56,020 –> 00:08:57,669 Hence, we would be interested

345 00:08:57,679 –> 00:08:59,030 in the distance between the

346 00:08:59,039 –> 00:09:00,630 limit point B and the

347 00:09:00,640 –> 00:09:01,919 sequence X and K

348 00:09:02,299 –> 00:09:03,960 simply because if the sequence

349 00:09:03,969 –> 00:09:05,309 was convergent, this would

350 00:09:05,320 –> 00:09:06,210 go to zero.

351 00:09:06,700 –> 00:09:08,010 However, the only distance

352 00:09:08,020 –> 00:09:09,489 we know at this point is

353 00:09:09,500 –> 00:09:11,429 the distance between A and

354 00:09:11,440 –> 00:09:11,989 XN.

355 00:09:12,489 –> 00:09:13,979 And using the subsequence,

356 00:09:13,989 –> 00:09:15,840 we know this is larger than

357 00:09:15,849 –> 00:09:16,510 NK.

358 00:09:17,030 –> 00:09:18,309 Then on the right hand side,

359 00:09:18,320 –> 00:09:20,039 we simply can use our triangle

360 00:09:20,049 –> 00:09:20,849 inequality.

361 00:09:21,650 –> 00:09:23,539 We just put B in the middle,

362 00:09:24,330 –> 00:09:25,869 then bring the one part on

363 00:09:25,880 –> 00:09:27,070 the other side and we get

364 00:09:27,080 –> 00:09:28,510 the following inequality.

365 00:09:28,909 –> 00:09:30,590 So we conclude the distance

366 00:09:30,599 –> 00:09:32,580 between any point B and a

367 00:09:32,590 –> 00:09:34,489 sequence number X and K is

368 00:09:34,500 –> 00:09:35,950 always larger or

369 00:09:35,960 –> 00:09:37,789 equal than this part here.

370 00:09:38,340 –> 00:09:40,190 However, seeing the NK here,

371 00:09:40,200 –> 00:09:41,590 you know, you can make the

372 00:09:41,599 –> 00:09:43,429 left hand side as large as

373 00:09:43,440 –> 00:09:43,950 you want.

374 00:09:44,669 –> 00:09:46,049 Hence, our conclusion is

375 00:09:46,059 –> 00:09:47,460 no matter which point B you

376 00:09:47,469 –> 00:09:48,070 choose.

377 00:09:48,080 –> 00:09:49,809 If you send K to infinity,

378 00:09:49,820 –> 00:09:51,650 this will never go to zero.

379 00:09:52,539 –> 00:09:53,969 So there can’t be any limit

380 00:09:53,979 –> 00:09:55,479 point for the subsequence.

381 00:09:55,979 –> 00:09:57,570 If you fix a point B here,

382 00:09:57,580 –> 00:09:58,770 you will always run away

383 00:09:58,780 –> 00:09:59,169 from it.

384 00:10:00,010 –> 00:10:01,440 And therefore A is not

385 00:10:01,450 –> 00:10:03,190 compact because we found

386 00:10:03,200 –> 00:10:04,570 a sequence which does not

387 00:10:04,580 –> 00:10:06,510 own any convergent subsequence.

388 00:10:07,169 –> 00:10:08,580 And by contras position,

389 00:10:08,590 –> 00:10:09,919 we finished our proof, a

390 00:10:09,929 –> 00:10:11,750 compact set is always

391 00:10:11,760 –> 00:10:12,390 bounded.

392 00:10:12,710 –> 00:10:13,109 OK.

393 00:10:13,119 –> 00:10:14,429 So now you know the notion

394 00:10:14,440 –> 00:10:16,260 compact for metric spaces

395 00:10:16,270 –> 00:10:17,520 and in the next videos, we

396 00:10:17,530 –> 00:10:18,809 will extend this to

397 00:10:18,820 –> 00:10:19,640 operators.

398 00:10:20,229 –> 00:10:21,609 So I hope I see you there.

399 00:10:21,679 –> 00:10:23,140 Thanks for listening and

400 00:10:23,150 –> 00:10:23,750 bye.

Q1: Let $(X,d)$ be a metric space. What is correct for a sequentially compact set $A \subseteq X$? Here $(x_n)_{n \in \mathbb{N}}$ denotes a sequence in $X$.

A1: If $x_n \in A$ for all $n \in \mathbb{N}$, then there is a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ that is convergent.

A2: There is a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ that is convergent.

A3: $(x_n)$ is a convergent sequence.

A4: Every subsequence of $(x_n)$ converges to an element in $A$.

A5: $A$ is unbounded.

Q2: Consider the metric space $(\mathbb{R}, d_{\mathrm{eucl.}})$. Which of the following sets is compact?

A1: ${0,1,2,3,4,5,6,7,8}$

A2: $\mathbb{Z}$

A3: $[0,1)$

A4: $[0,\infty)$

Q3: Consider the metric space $(\mathbb{R}, d_{\mathrm{discr.}})$. Which of the following sets is compact?

A1: ${0,1,2,3,4,5,6,7,8}$

A2: $\mathbb{Z}$

A3: $[0,1]$

A4: $(0,1)$

A5: $[0,\infty)$

Q4: Consider the metric space $(\mathbb{R}, d_{\mathrm{discr.}})$. Which statement is correct?

A1: $A$ compact $\Rightarrow$ $A$ closed and bounded

A2: $A$ closed and bounded $\Rightarrow$ $A$ compact

A3: $[0,1]$ is not open

A4: $(0,1)$ is not closed

A5: $[0,\infty)$ is not bounded